1. Coulomb's law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
2. The electric field intensity is defined as the force per unit charge. The electric field intensity due to a single point charge is calculated using Coulomb's law. For multiple point charges, the total electric field is calculated using the principle of superposition.
3. Continuous charge distributions can also produce electric fields. These include line charge distributions with linear charge density, surface charge distributions with surface charge density, and volume charge distributions with volume charge density. The electric field due to a uniform line charge distribution can be calculated by treating
The document discusses deformation spectra for single-degree-of-freedom (SDF) linear systems subjected to base excitation. It presents the equations of motion for an SDF system with a moving base and defines terms like relative displacement and pseudo-acceleration. Graphs of deformation spectra are shown for half-cycle acceleration and velocity pulses. Key aspects of the spectra under different inputs are described, including asymptotic behavior and sensitivity to displacement, velocity, and acceleration portions of the input.
This document contains lecture material on moments and couples from a engineering mechanics course. It includes:
- Definitions of moment as the turning effect of a force and how to calculate moment magnitude using the perpendicular distance from the axis (moment arm).
- Details on sign conventions and using the right hand rule to determine moment direction.
- An explanation of Varignon's theorem which states the sum of all individual moments equals the moment of the resultant force.
- Examples of calculating moments using scalar and vector methods, as well as resolving force systems into equivalent force-couple systems.
- The definition of a couple as two equal and opposite parallel forces separated by a perpendicular distance, and characteristics such as couple
This document provides an overview of dry friction and Coulomb's theory of dry friction. It introduces key concepts such as static and kinetic friction, coefficients of friction, and limitations of Coulomb's model. Example problems are provided to demonstrate applications of dry friction analysis to determine equilibrium, potential for sliding or tipping, and minimum horizontal forces required to move objects. The document is from a engineering mechanics course focusing on rigid body analysis involving dry frictional forces.
The document discusses the concepts of resultants and distributed forces in engineering mechanics. It covers determining the magnitude and location of the resultant for parallel, non-parallel and distributed force systems. Methods are presented for calculating the resultant and locating its line of action for uniform and trapezoidal distributed loadings on objects. Examples problems are also provided and worked through.
6161103 4.10 reduction of a simple distributed loadingetcenterrbru
1) The document describes how to reduce a distributed loading over a surface into an equivalent single resultant force.
2) The magnitude of the resultant force is equal to the total area under the loading diagram and its line of action passes through the centroid of this area.
3) Examples are provided to demonstrate calculating the magnitude and location of the equivalent resultant force for different loading diagrams including rectangular, triangular, and trapezoidal shapes.
The document contains multiple conceptual physics problems involving conservation of energy.
1) A system of two cylinders connected by a cord over a frictionless peg is released from rest. The system's mechanical energy is conserved, so the potential energy decreases and kinetic energy increases.
2) Estimation problems calculate the minimum time to climb stairs or the Empire State Building using assumptions about maximum metabolic rate and efficiency.
3) Multiple other problems apply conservation of mechanical energy to calculate quantities like tension, equilibrium angles, or speeds in various physical systems like swings or circular motion.
This document summarizes the key concepts around harmonic excitation of undamped and damped systems from Chapter 2. It introduces the important concept of resonance that occurs when the driving frequency matches the natural frequency of the system. For an undamped system under harmonic excitation, the response is the sum of the homogeneous and particular solutions. The particular solution assumes the form of the driving force. When the driving frequency approaches the natural frequency, the amplitude of the response increases dramatically. For a damped system, the particular solution includes a phase shift.
The document discusses deformation spectra for single-degree-of-freedom (SDF) linear systems subjected to base excitation. It presents the equations of motion for an SDF system with a moving base and defines terms like relative displacement and pseudo-acceleration. Graphs of deformation spectra are shown for half-cycle acceleration and velocity pulses. Key aspects of the spectra under different inputs are described, including asymptotic behavior and sensitivity to displacement, velocity, and acceleration portions of the input.
This document contains lecture material on moments and couples from a engineering mechanics course. It includes:
- Definitions of moment as the turning effect of a force and how to calculate moment magnitude using the perpendicular distance from the axis (moment arm).
- Details on sign conventions and using the right hand rule to determine moment direction.
- An explanation of Varignon's theorem which states the sum of all individual moments equals the moment of the resultant force.
- Examples of calculating moments using scalar and vector methods, as well as resolving force systems into equivalent force-couple systems.
- The definition of a couple as two equal and opposite parallel forces separated by a perpendicular distance, and characteristics such as couple
This document provides an overview of dry friction and Coulomb's theory of dry friction. It introduces key concepts such as static and kinetic friction, coefficients of friction, and limitations of Coulomb's model. Example problems are provided to demonstrate applications of dry friction analysis to determine equilibrium, potential for sliding or tipping, and minimum horizontal forces required to move objects. The document is from a engineering mechanics course focusing on rigid body analysis involving dry frictional forces.
The document discusses the concepts of resultants and distributed forces in engineering mechanics. It covers determining the magnitude and location of the resultant for parallel, non-parallel and distributed force systems. Methods are presented for calculating the resultant and locating its line of action for uniform and trapezoidal distributed loadings on objects. Examples problems are also provided and worked through.
6161103 4.10 reduction of a simple distributed loadingetcenterrbru
1) The document describes how to reduce a distributed loading over a surface into an equivalent single resultant force.
2) The magnitude of the resultant force is equal to the total area under the loading diagram and its line of action passes through the centroid of this area.
3) Examples are provided to demonstrate calculating the magnitude and location of the equivalent resultant force for different loading diagrams including rectangular, triangular, and trapezoidal shapes.
The document contains multiple conceptual physics problems involving conservation of energy.
1) A system of two cylinders connected by a cord over a frictionless peg is released from rest. The system's mechanical energy is conserved, so the potential energy decreases and kinetic energy increases.
2) Estimation problems calculate the minimum time to climb stairs or the Empire State Building using assumptions about maximum metabolic rate and efficiency.
3) Multiple other problems apply conservation of mechanical energy to calculate quantities like tension, equilibrium angles, or speeds in various physical systems like swings or circular motion.
This document summarizes the key concepts around harmonic excitation of undamped and damped systems from Chapter 2. It introduces the important concept of resonance that occurs when the driving frequency matches the natural frequency of the system. For an undamped system under harmonic excitation, the response is the sum of the homogeneous and particular solutions. The particular solution assumes the form of the driving force. When the driving frequency approaches the natural frequency, the amplitude of the response increases dramatically. For a damped system, the particular solution includes a phase shift.
The document provides conceptual problems and their solutions related to Newton's Laws of motion.
1) A problem asks how to determine if a limousine is changing speed or direction using a small object on a string. The solution is that if the string remains vertical, the reference frame is inertial.
2) Another problem asks for two situations where apparent weight in an elevator is greater than true weight. The solution states this occurs when the elevator accelerates upward, either slowing down or speeding up.
3) A third problem involves forces between blocks and identifies which constitute Newton's third law pairs. The normal forces between blocks and between a block and table are identified as third law pairs.
1. The document contains solutions to physics problems involving Coulomb's law and electric fields. Problems calculate the number of electrons in a charge, magnitude of electric forces between charges, electric field strength created by charges, and acceleration of charges in electric fields.
2. Key steps shown include using Coulomb's law to calculate electric forces and equating force to charge times electric field to determine field strength.
3. Charges, distances, fields, and forces are accounted for with positive and negative signs to determine direction according to the conventions of electric field vectors and forces between charges.
The document provides lecture material on the equilibrium of coplanar and non-coplanar force systems. It includes objectives, definitions, conditions, procedures, and examples for determining unknown forces and reactions in static equilibrium systems. Graphical methods like the triangle and polygon laws and Lami's theorem are presented. Several practice problems and their solutions are provided to illustrate the application of the equilibrium equations to concurrent, parallel, and non-parallel/non-concurrent force systems. Homework assignments consisting of additional practice problems are also given.
1) The document discusses Castigliano's theorems, which relate the partial derivative of a structure's strain energy to its deflections under applied forces or moments.
2) Castigliano's first theorem states that the partial derivative of strain energy with respect to an applied force is equal to the deflection of the point of application in the direction of the force.
3) Examples are provided to demonstrate calculating deflections of beams under various load conditions using Castigliano's first theorem.
1) A point charge moving through an electric field is shown to follow a parabolic trajectory given by z = -1.5×1010t2 m. At t = 3 μs, its position is found to be P(0.90, 0, -0.135) m.
2) A point charge moving through a uniform magnetic field follows a circular path. The equations of motion are derived and solved, giving the position, velocity, and kinetic energy at t = 3 μs.
3) Forces on current loops and filaments in various magnetic field configurations are calculated.
6161103 4.9 further reduction of a force and couple systemetcenterrbru
The document describes methods to reduce systems of forces and moments acting on a rigid body to an equivalent single resultant force and moment. It discusses reducing force systems that are concurrent, coplanar, or parallel to a single resultant force located at a point P. The location of P can be determined by satisfying the moment condition such that the resultant force and moment are perpendicular. Examples demonstrate applying these methods to solve for the magnitude, direction, and application point of the equivalent resultant force.
This document discusses Castigliano's theorem for calculating deflections in elastic structures. Castigliano's theorem states that the deflection of a point on a structure due to a load is equal to the partial derivative of the total strain energy with respect to that load. The document provides examples of using Castigliano's theorem to calculate deflections for structures undergoing bending, shear, tension, and compression. It also discusses using a "dummy load" and setting the partial derivative of energy to the load equal to zero to solve for an unknown force.
Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.
Here are the key differences between a particle and a rigid body in mechanics:
Particle:
- Has no size or internal structure, it is considered a point object.
- Cannot transfer or support moments/torques. Only forces can act on a particle.
Rigid Body:
- Has size, shape and internal structure. It is an extended object.
- Can transfer and support both forces and moments/torques at its different points.
Other differences:
- Equations of equilibrium for a particle involve only forces. Equations for a rigid body involve both forces and moments.
- Deformations are not considered for a particle as it has no internal structure. Deformations may need to be
1. This document provides solutions to physics problems involving magnetic fields and forces on currents and charges in magnetic fields. It solves 30 problems involving calculating magnetic fields, forces on currents and charges, and directions of forces. Diagrams are included to illustrate some of the problems and solutions.
2. The problems calculate things like maximum forces on wires in magnetic fields, directions of forces on electrons, magnetic fields produced by currents, radii of circular electron paths, and effects of magnetic fields on compasses.
3. Right-hand rules are used to determine directions of magnetic fields and forces on charges. Formulas for magnetic fields and forces are applied to calculate values needed to solve the problems.
The document contains 10 problems involving electromagnetic induction and Maxwell's equations. Problem 10.1 involves calculating the voltage and current in a circuit with a changing magnetic flux. Problem 10.2 replaces a voltmeter with a resistor and calculates the resulting current. Problem 10.3 calculates the emf induced in closed paths with changing magnetic fluxes.
This document contains physics formulas and concepts related to waves, electricity, electromagnetism, electronics, and radioactivity. It includes equations for oscillation, wavelength, interference, charge, current, potential difference, resistance, electromotive force, transformers, alpha decay, beta decay, gamma emission, and nuclear energy. The document provides definitions and explanations for key terms as well as examples of applying the formulas.
1) A student pointed out that a sailboat with a fan blowing into its sails could in fact move forward, because the air molecules bouncing off the sail would impart twice the momentum change experienced going through the fan, providing a net forward force.
2) A 2000-kg car traveling at 30 m/s collided perfectly inelastically with another 2000-kg car traveling at 10 m/s, sticking together with a final speed of 20 m/s. 20% of the initial kinetic energy was lost to heat and deformation.
3) A 16-g bullet fired into a 1.5-kg ballistic pendulum was calculated to have a speed of 45 m/s before impact by applying
Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.
This document discusses Castigliano's theorems for analyzing stresses and strains in structures. It explains that Castigliano's first theorem states that the partial derivative of a structure's strain energy with respect to an applied force equals the displacement at the point of application of that force. Castigliano's second theorem states that the partial derivative of strain energy with respect to a displacement equals the force that produces that displacement. The document provides mathematical expressions to calculate strain energy and uses these theorems to analyze beam deflections under applied loads.
This document summarizes key concepts about electric potential and potential energy from Chapter 3:
1) Electric potential is defined as the work required per unit charge to move a test charge between two points in an electric field, similar to how gravitational potential is defined. 2) In a uniform electric field, the electric potential decreases as one moves in the direction of the field lines, corresponding to a decrease in potential energy for a positive charge. 3) The change in electric potential between two points depends only on the endpoints and not the path taken, as electric fields are conservative.
Principles of soil dynamics 3rd edition das solutions manualHuman2379
Full download: https://goo.gl/MyzREj
principles of soil dynamics pdf
soil dynamics and liquefaction
fundamentals of soil dynamics and earthquake engineering
principles of foundation engineering
Strain energy is a type of potential energy that is stored in a structural member as a result of elastic deformation. The external work done on such a member when it is deformed from its unstressed state is transformed into (and considered equal to the strain energy stored in it.
This document discusses multi-degree-of-freedom (MDOF) systems and their analysis. It introduces concepts such as flexibility and stiffness matrices, natural frequencies and mode shapes, orthogonality of modes, and equations of motion. Methods for analyzing free and forced vibration of MDOF systems in the time domain are presented, including modal superposition and direct integration. An example 3DOF system is analyzed to illustrate the concepts.
EMF ELECTROSTATICS:
Coulomb’s Law, Electric Field of Different Charge Configurations using Coulomb’s Law, Electric Flux, Field Lines, Gauss’s Law in terms of E (Integral Form and Point Form), Applications of Gauss’s Law, Curl of the Electric Field, Electric Potential, Calculation of Electric Field Through Electric Potential for given Charge Configuration, Potential Gradient, The Dipole, Energy density in the Electric field.
1. Static electricity can be demonstrated by rubbing a glass rod with silk and hanging it from a thread. A second glass rod rubbed with silk will repel the first, while a hard rubber rod rubbed with fur will attract the glass rod.
2. The force between two electrically charged bodies follows an inverse square law, where the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
3. Electric potential is defined as the work required to move a unit positive charge from a reference point (usually infinity) to the point where the potential is being measured, without producing an acceleration.
The document provides conceptual problems and their solutions related to Newton's Laws of motion.
1) A problem asks how to determine if a limousine is changing speed or direction using a small object on a string. The solution is that if the string remains vertical, the reference frame is inertial.
2) Another problem asks for two situations where apparent weight in an elevator is greater than true weight. The solution states this occurs when the elevator accelerates upward, either slowing down or speeding up.
3) A third problem involves forces between blocks and identifies which constitute Newton's third law pairs. The normal forces between blocks and between a block and table are identified as third law pairs.
1. The document contains solutions to physics problems involving Coulomb's law and electric fields. Problems calculate the number of electrons in a charge, magnitude of electric forces between charges, electric field strength created by charges, and acceleration of charges in electric fields.
2. Key steps shown include using Coulomb's law to calculate electric forces and equating force to charge times electric field to determine field strength.
3. Charges, distances, fields, and forces are accounted for with positive and negative signs to determine direction according to the conventions of electric field vectors and forces between charges.
The document provides lecture material on the equilibrium of coplanar and non-coplanar force systems. It includes objectives, definitions, conditions, procedures, and examples for determining unknown forces and reactions in static equilibrium systems. Graphical methods like the triangle and polygon laws and Lami's theorem are presented. Several practice problems and their solutions are provided to illustrate the application of the equilibrium equations to concurrent, parallel, and non-parallel/non-concurrent force systems. Homework assignments consisting of additional practice problems are also given.
1) The document discusses Castigliano's theorems, which relate the partial derivative of a structure's strain energy to its deflections under applied forces or moments.
2) Castigliano's first theorem states that the partial derivative of strain energy with respect to an applied force is equal to the deflection of the point of application in the direction of the force.
3) Examples are provided to demonstrate calculating deflections of beams under various load conditions using Castigliano's first theorem.
1) A point charge moving through an electric field is shown to follow a parabolic trajectory given by z = -1.5×1010t2 m. At t = 3 μs, its position is found to be P(0.90, 0, -0.135) m.
2) A point charge moving through a uniform magnetic field follows a circular path. The equations of motion are derived and solved, giving the position, velocity, and kinetic energy at t = 3 μs.
3) Forces on current loops and filaments in various magnetic field configurations are calculated.
6161103 4.9 further reduction of a force and couple systemetcenterrbru
The document describes methods to reduce systems of forces and moments acting on a rigid body to an equivalent single resultant force and moment. It discusses reducing force systems that are concurrent, coplanar, or parallel to a single resultant force located at a point P. The location of P can be determined by satisfying the moment condition such that the resultant force and moment are perpendicular. Examples demonstrate applying these methods to solve for the magnitude, direction, and application point of the equivalent resultant force.
This document discusses Castigliano's theorem for calculating deflections in elastic structures. Castigliano's theorem states that the deflection of a point on a structure due to a load is equal to the partial derivative of the total strain energy with respect to that load. The document provides examples of using Castigliano's theorem to calculate deflections for structures undergoing bending, shear, tension, and compression. It also discusses using a "dummy load" and setting the partial derivative of energy to the load equal to zero to solve for an unknown force.
Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.
Here are the key differences between a particle and a rigid body in mechanics:
Particle:
- Has no size or internal structure, it is considered a point object.
- Cannot transfer or support moments/torques. Only forces can act on a particle.
Rigid Body:
- Has size, shape and internal structure. It is an extended object.
- Can transfer and support both forces and moments/torques at its different points.
Other differences:
- Equations of equilibrium for a particle involve only forces. Equations for a rigid body involve both forces and moments.
- Deformations are not considered for a particle as it has no internal structure. Deformations may need to be
1. This document provides solutions to physics problems involving magnetic fields and forces on currents and charges in magnetic fields. It solves 30 problems involving calculating magnetic fields, forces on currents and charges, and directions of forces. Diagrams are included to illustrate some of the problems and solutions.
2. The problems calculate things like maximum forces on wires in magnetic fields, directions of forces on electrons, magnetic fields produced by currents, radii of circular electron paths, and effects of magnetic fields on compasses.
3. Right-hand rules are used to determine directions of magnetic fields and forces on charges. Formulas for magnetic fields and forces are applied to calculate values needed to solve the problems.
The document contains 10 problems involving electromagnetic induction and Maxwell's equations. Problem 10.1 involves calculating the voltage and current in a circuit with a changing magnetic flux. Problem 10.2 replaces a voltmeter with a resistor and calculates the resulting current. Problem 10.3 calculates the emf induced in closed paths with changing magnetic fluxes.
This document contains physics formulas and concepts related to waves, electricity, electromagnetism, electronics, and radioactivity. It includes equations for oscillation, wavelength, interference, charge, current, potential difference, resistance, electromotive force, transformers, alpha decay, beta decay, gamma emission, and nuclear energy. The document provides definitions and explanations for key terms as well as examples of applying the formulas.
1) A student pointed out that a sailboat with a fan blowing into its sails could in fact move forward, because the air molecules bouncing off the sail would impart twice the momentum change experienced going through the fan, providing a net forward force.
2) A 2000-kg car traveling at 30 m/s collided perfectly inelastically with another 2000-kg car traveling at 10 m/s, sticking together with a final speed of 20 m/s. 20% of the initial kinetic energy was lost to heat and deformation.
3) A 16-g bullet fired into a 1.5-kg ballistic pendulum was calculated to have a speed of 45 m/s before impact by applying
Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.
This document discusses Castigliano's theorems for analyzing stresses and strains in structures. It explains that Castigliano's first theorem states that the partial derivative of a structure's strain energy with respect to an applied force equals the displacement at the point of application of that force. Castigliano's second theorem states that the partial derivative of strain energy with respect to a displacement equals the force that produces that displacement. The document provides mathematical expressions to calculate strain energy and uses these theorems to analyze beam deflections under applied loads.
This document summarizes key concepts about electric potential and potential energy from Chapter 3:
1) Electric potential is defined as the work required per unit charge to move a test charge between two points in an electric field, similar to how gravitational potential is defined. 2) In a uniform electric field, the electric potential decreases as one moves in the direction of the field lines, corresponding to a decrease in potential energy for a positive charge. 3) The change in electric potential between two points depends only on the endpoints and not the path taken, as electric fields are conservative.
Principles of soil dynamics 3rd edition das solutions manualHuman2379
Full download: https://goo.gl/MyzREj
principles of soil dynamics pdf
soil dynamics and liquefaction
fundamentals of soil dynamics and earthquake engineering
principles of foundation engineering
Strain energy is a type of potential energy that is stored in a structural member as a result of elastic deformation. The external work done on such a member when it is deformed from its unstressed state is transformed into (and considered equal to the strain energy stored in it.
This document discusses multi-degree-of-freedom (MDOF) systems and their analysis. It introduces concepts such as flexibility and stiffness matrices, natural frequencies and mode shapes, orthogonality of modes, and equations of motion. Methods for analyzing free and forced vibration of MDOF systems in the time domain are presented, including modal superposition and direct integration. An example 3DOF system is analyzed to illustrate the concepts.
EMF ELECTROSTATICS:
Coulomb’s Law, Electric Field of Different Charge Configurations using Coulomb’s Law, Electric Flux, Field Lines, Gauss’s Law in terms of E (Integral Form and Point Form), Applications of Gauss’s Law, Curl of the Electric Field, Electric Potential, Calculation of Electric Field Through Electric Potential for given Charge Configuration, Potential Gradient, The Dipole, Energy density in the Electric field.
1. Static electricity can be demonstrated by rubbing a glass rod with silk and hanging it from a thread. A second glass rod rubbed with silk will repel the first, while a hard rubber rod rubbed with fur will attract the glass rod.
2. The force between two electrically charged bodies follows an inverse square law, where the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
3. Electric potential is defined as the work required to move a unit positive charge from a reference point (usually infinity) to the point where the potential is being measured, without producing an acceleration.
Gen Phy 2 Q1L3 Electric Charge and Coulumb's Law.pptxJeffrey Alemania
* Given: q = 8.00 x 10-9 C
* Side of cube = 0.200 m
* Distance between charge and side = 0.141 m
* Area of each face = 0.200 x 0.200 = 0.0400 m2
* Using Gauss's law: φE = EA
* E = kq/r2 = (9 x 109 Nm2/C2)(8.00 x 10-9 C)/(0.141 m)2 = 4.80 x 105 N/C
* φE = (4.80 x 105 N/C)(0.0400 m2) = 1.92 x 10-3 Nm2/C
Therefore
1) The document discusses electric field intensity due to various charge distributions including point charges, line charges, and charge distributions on surfaces.
2) It provides examples calculating the electric field intensity at given points due to single or multiple point charges using the inverse square law expression for a point charge.
3) Another example calculates the electric field intensity at the center of an equilateral triangle formed by three line charges of different linear charge densities arranged along the sides.
This document provides a summary of a physics exam paper containing multiple choice and numerical questions. It tests concepts in circuits, optics, mechanics, and other areas of physics. The summary focuses on highlighting the key topics covered and question formats rather than reproducing content.
The paper contains two sections - the first with multiple choice questions testing circuits, optics, waves, and electrostatics concepts. Answers can have one or more options correct. The second section involves numerical questions where the answer is a single integer between 0-9. Question formats include experimental analysis, mechanics problems, and converting between electrical measuring devices.
Electricity is associated with the presence and motion of electric charge. There are two types of electricity: static electricity and current electricity. Static electricity results from an imbalance of negative and positive charges in an object that can build up until being discharged. Electric charge is measured in coulombs and there are two types: positive and negative.
The electric field is the region of space surrounding an electrically charged object where an electric force can be detected. It is represented by electric field lines. The electric field intensity is the electric force per unit charge and is measured in newtons per coulomb. Coulomb's law describes the electric force between two point charges. Gauss's law relates the electric flux through a closed surface to the net
1. The document provides the marking scheme for a Physics exam with 15 multiple choice and numerical questions.
2. Key concepts assessed include photoelectric effect, electromagnetic waves, optics, electricity and magnetism.
3. Formulas, concepts, and steps are provided for full and partial credit on questions.
Lecture Notes: EEEC6430310 Electromagnetic Fields And Waves - Transmission LineAIMST University
This document discusses electromagnetic waves and transmission lines. It begins by introducing electromagnetic waves and how they propagate at the speed of light. Transmission lines guide electromagnetic waves from one place to another. The document then discusses transverse electromagnetic waves on transmission lines and how their characteristic impedance is determined. It provides examples of calculating properties of transmission lines like coaxial cable. Overall, the document provides an introduction to transmission lines and how they propagate electromagnetic waves through circuit analysis and wave equations.
This document discusses electromagnetic fields and waves. It contains examples and explanations of the electric field intensity due to point charges located at both the origin and other positions in space. It also discusses the electric field due to continuous charge distributions on surfaces and lines. Gauss's law is introduced, which relates the electric flux through a closed surface to the enclosed charge. Examples are provided to demonstrate how Gauss's law can be used to determine electric fields of symmetric charge configurations.
Physics; presentation electrostat; -harsh kumar;- xii science; -roll no 08Harsh Kumar
This presentation summarizes key concepts in electrostatics, including:
1. Electrostatics is the study of electrical properties of systems with charges at rest. Charge is a fundamental property that explains electrical behavior and can be positive or negative.
2. Coulomb's law describes the attractive force between two point charges, directly proportional to the product of charges and inversely proportional to the square of the distance between them.
3. Electric field intensity is defined as the force experienced by a unit positive charge at a point in an electric field. Gauss's law relates the electric flux through a closed surface to the net charge enclosed.
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3. The electric field intensity is defined as the electrostatic force per unit positive test charge. It depends on the source charge and distance from the source charge. According
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2. Frictional electricity is produced by rubbing two materials together, which causes the transfer of electrons between the materials. Coulomb's law describes the electrostatic force between two point charges.
3. Continuous charge distribution can be linear, surface, or volume charge densities that describe the charge per unit length, area, or volume respectively within that distribution.
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3. Frictional electricity is produced by rubbing two materials together, causing a transfer of electrons between them. This leaves one material positively charged and the other negatively charged.
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Emf unit i
1. UNIT - I
ELECTROSTATICS
Introduction
Charges at rest produce Static Electric Field or Electrostatic field.
Field: It is the existing space in particular area due to some elements.
1.1) Coulomb’s Law
Coulomb stated that the force between two point charges separated in a vacuum or free space
by a distance which is large compared to their size is
(i) proportional to the charge on each
(ii) inversely proportional to the square of the distance between them
(iii) directed along the line joining the charges
ߙ ܨ
ܳଵܳଶ
ܴଶ
ଵଶ
Fig. 1.1 If Q1 and Q2 have like signs the vector force ࡲ on Q2 is in the same direction as ࡾ
ܨ = ݇
ܳଵܳଶ
ܴଶ
ଵଶ
(1.1)
Where Q1 and Q2 are the positive or negative quantities of charge, R12 is the separation, and k
is proportionality constant. If the International System of Units (SI) is used. In SI units,
charges Q1 and Q2 are in coulombs (C), the distance R12 is in meters (m), and the force F is in
newtons (N) so that k = 1/4πεo. The constant εo is known as the permittivity of free space (in
farads per meter) and has the value
εo= 8.854×10-12
≈
ଵషవ
ଷగ
F/m
k = 1/(4πεo) = 9 × 109
m/F
Thus Eq. (1.1) becomes
2. ܨ =
1
4ߨߝ
ܳଵܳଶ
ܴଶ
ଵଶ
(1.2)
If point charges Q1 and Q2 are located at points having position vectors ࢘ and ࢘, then the
force F2 on Q1 due to Q2, shown in Figure 1.1, is given by
ࡲ =
1
4ߨߝ
ܳଵܳଶ
ܴଶ
ଵଶ
ࢇࡾ
(1.3)
Where
ࢇࡾ
=
ࡾ
|ࡾ|
(1.4)
ࡾ = ࢘ − ࢘
ܴଵଶ = |ࡾ|
By substituting eq. (1.4) into eq. (1.3), we may write eq. (1.3) as
ࡲ =
1
4ߨߝ
ܳଵܳଶ
ܴଷ
ଵଶ
ࡾ (1.5)
It is worthwhile to note that
The force ࡲ on Q1 due to Q2 is given by
ࡲ = |ࡲ|ࢇࡾ
= |ࡲ|(−ࢇࡾ
) = −ࡲ
ࡲ = −ࡲ (1.6)
Problems
1. The charge Q2 = 10 µc is located at (3,1,0) and Q1 = 50 µc is located at (-1,1,-3). Find
the force on Q1.
Ans. R21 = (-1-3) i + (1-1) j + (-3-0) k = -4i-3k
| R21| = ඥ(−4)ଶ + (−3)ଶ = 5
ࡲ =
1
4ߨߝ
ܳଵܳଶ
ܴଷ
ଶଵ
ࡾ
ࡲ =
9 × 10ଽ
10 × 10ି
× 50 × 10ି
5ଷ
(−4i − 3k)
ࡲ = −. − . ૡ
2. A point charge Q1 = 300 µc located at (1,-1,-3) experiences a force F1= 8i-8j+4k due
to a point charge Q2 at (3,-3,-2)m. Determine Q2.
Ans. R21 = (1-3) i + (3-1) j + (2-3) k = -2 i +2 j - k
3. | R21| = ඥ(−2)ଶ + (2)ଶ + (−1)ଶ = 3
ࡲ =
1
4ߨߝ
ܳଵܳଶ
ܴଷ
ଶଵ
ࡾ
(8i − 8j + 4k ) =
9 × 10ଽ
ܳଶ × 300 × 10ି
3ଷ
(−2 i + 2 j − k)
Q2 = -40 µc.
1.2) Force due to N no. of charges:
If we have more than two point charges, we can use the principle of superposition to
determine the force on a particular charge. The principle states that if there are N charges Q1,
Q2, Q3..................... QN located, respectively, at points with position vectors
࢘, ࢘, ࢘ … … … … … ࢘ࡺ , the resultant force ࡲ on a charge Q located at point(p) ࢘ is the
vector sum of the forces exerted on Q by each of the charges Q1, Q2, Q3..................... QN. Hence:
ࡲ = ࡲ + ࡲ + ⋯ + ࡲࡺ
ࡲ =
1
4ߨߝ
ܳܳଶ
ܴଷ
ଵ
ࡾ +
1
4ߨߝ
ܳܳଶ
ܴଷ
ଶ
ࡾ + ⋯ +
1
4ߨߝ
ܳܳଶ
ܴଷ
ே
ࡾࡺ (1.7)
ࡲ =
ܳ
4ߨߝ
ܳ
ܴଷ
ࡾ
ே
ୀଵ
(1.8)
ࡲ = 9 × 10ଽ
ܳ
ܳ
ܴଷ
ࡾ
ே
ୀଵ
(1.9)
Problems
1. Find the force on a 100 µc charge at (0,0,3)m. If four like charges of 20 µc are located
on x and y axes at ±4m.
Ans. R1p = (0-4) i + (0-0) j + (3-0) k = -4 i +3 k
| R1p| = ඥ(−4)ଶ + (3)ଶ = 5
R2p = (0-0) i + (0-4) j + (3-0) k = -4 j +3 k
| R2p| = ඥ(−4)ଶ + (3)ଶ = 5
R3p = (0+4) i + (0-0) j + (3-0) k = 4 i +3 k
4. | R3p| = ඥ(4)ଶ + (3)ଶ = 5
R4p = (0-0) i + (0+4) j + (3-0) k = 4 i +3 k
| R4p| = ඥ(4)ଶ + (3)ଶ = 5
ࡲ = 9 × 10ଽ
ܳ
ܳ
ܴଷ
ࡾ
ே
ୀଵ
ࡲ =
9 × 10ଽ
100 × 10ି
× 20 × 10ି
5ଷ
(−4 i + 3 k − 4 i + 3 k + 4 i + 3 k + 4 i + 3 k)
ࡲ = . ૠ ࡷ ܰ
2. Two small diameter 10 gm dielectric balls can slide freely on a vertical plastic channel. Each
ball carries a negative charge of 1 nc. Find the separation between the balls if the upper ball is
restrained from moving.
Ans. Gravitational force and coulomb force must be equal
F2 = mg = 10 × 10-3
× 9.8 = 9.8× 10-2
N.
ܨଵ =
1
4ߨߝ
ܳܳଶ
ܴଶ
ଵ
Therefore F1 = F2
9.8× 10-2
=
ଽ×ଵవ (ଵ× ଵషల)మ
ோమ
R=0.302 m
1.3) Electric Field Intensity or Electric Field Strength (ࡱ):
It is the force per unit charge when placed in electric field.
Fig 1.2 The lines of force due to a pair of charges, one positive and the other negative
5. Fig 1.3 The lines of force due to a pair of positive charges
An electric field is said to exist if a test charge kept in the space surrounding another charge,
then it will experience a force.
Thus,
ࡱ = lim
ொ→
ࡲ࢚
ܳ௧
(1.10)
or simply
ࡱ =
ࡲ࢚
ܳ௧
(1.11)
The electric field intensity ࡱ is obviously in the direction of the force ࡲ and is measured in
newton/coulomb or volts/meter. The electric field intensity at point ࢚࢘ due to a point charge
located at ࢘ is readily obtained from eq. (1.3) as
ࡲ࢚ =
1
4ߨߝ
ܳଵܳ௧
ܴଶ
ଵ௧
ࢇࡾ࢚
ࡲ࢚
ܳ௧
=
1
4ߨߝ
ܳଵ
ܴଶ
ଵ௧
ࢇࡾ࢚
ࡱ =
1
4ߨߝ
ܳଵ
ܴଶ
ଵ௧
ࢇࡾ࢚
(1.12)
1.3.1) Electric field due to N no. of charges:
If we have more than two point charges, we can use the principle of superposition to
determine the force on a particular charge. The resultant force ࡲ on a charge Q located at
point(p) ࢘ is the vector sum of the forces exerted on Q by each of the charges Q1, Q2,
Q3..................... QN. Hence:
ࡲ = ࡲ + ࡲ + ⋯ + ࡲࡺ
From eq. (1.9)
6. ࡲ = 9 × 10ଽ
ܳ
ܳ
ܴଷ
ࡾ
ே
ୀଵ
We know that
ࡱ =
ࡲ
ܳ
ࡱ = 9 × 10ଽ
ܳ
ܴଷ
ࡾ
ே
ୀଵ
(1.13)
1.3.2) Electric Fields Due to Continuous Charge Distributions:
So far we have only considered forces and electric fields due to point charges, which
are essentially charges occupying very small physical space. It is also possible to have
continuous charge distribution along a line, on a surface, or in a volume as illustrated in
figure 1.2.
Fig. 1.4 volume charge distribution and charge elements
It is customary to denote the line charge density, surface charge density, and volume
charge density by λ (in C/m), σ (in C/m2
), and ρv (in C/m3
), respectively.
Line charge density: when the charge is distributed over linear element, then the line charge
density is the charge per unit length.
λ = lim
ௗ→
݀ݍ
݈݀
where dq is the charge on a linear element dl.
Surface charge density: when the charge is distributed over surface, then the surface charge
density is the charge per unit area.
7. σ = lim
ௗ௦→
݀ݍ
݀ݏ
where dq is the charge on a surface element ds.
Volume charge density: when the charge is confined within a volume, then the volume
charge density is the charge per unit volume.
ρ୴
= lim
ௗ௩→
݀ݍ
݀ݒ
where dq is the charge contained in a volume element dv.
1.3.3) Electric field due to line charge:
Consider a uniformly charged wire of length L m, the charge being assumed to be
uniformly distributed at the rate of λ (linear charge density) c/m. Let P be any point at which
electric field intensity has to be determined.
Consider a small elemental length dx at a distance x meters from the left end
of the wire, the corresponding charge element is λ dx. Divide the wire into a large number of
such small elements, each element will render its contribution towards the production of field
at P.
Fig 1.5 Evaluation E due to line charge
Let dE be field due to the charge element λ dx. It has a component dEx along x-axis
and dEy along y-axis.
dE = dEx ax + dEy ay
8. we know that
݀ܧ =
λ dx
4ߨߝݎଶ
(1.14)
From figure 1.5 we can write,
dEx = dE cosߠ (1.15)
dEy = dE sinߠ (1.16)
Therefore we can write,
݀ܧ௫ =
λ cosθ dx
4ߨߝݎଶ
(1.17)
݀ܧ௬ =
λ sinθ dx
4ߨߝݎଶ
(1.18)
Substituting eqs. (1.17) and (1.18) in dE we get,
݀ࡱ =
λ cosθ dx
4ߨߝݎଶ
ࢇ࢞ +
λ sinθ dx
4ߨߝݎଶ
ࢇ࢟ (1.19)
From figure 1.5 we write
L1-x = h cotߠ (1.20)
-dx = -h cosec2
ߠ dߠ
(1.21)
r = h cosecߠ
(1.22)
Substituting equations (1.20), (1.21) and (1.22) in 1.19, we get
݀ࡱ =
λ cosθ dθ
4ߨߝℎ
ࢇ࢞ +
λ sinθ dθ
4ߨߝℎ
ࢇ࢟ (1.23)
The electric field intensity E due to whole length of the wire
ࡱ = න ݀ࡱ ݀ߠ
ࣂୀ࣊ିࢻ
ࣂୀࢻ
ࡱ = න
λ cosθ dθ
4ߨߝℎ
ࢇ࢞ +
λ sinθ dθ
4ߨߝℎ
ࢇ࢟ ൨ ݀ߠ
ࣂୀ࣊ିࢻ
ࣂୀࢻ
9. ࡱ =
ߣ
4ߨߝℎ
ൣࢇ ߠ݊݅ݏ࢞ − ܿࢇ ߠݏ࢟൧
ࢻ
࣊ିࢻ
ࡱ =
ߣ
4ߨߝℎ
ൣ(ߙ݊݅ݏଶ − ߙ݊݅ݏଵ)ࢇ࢞ + (ܿߙݏଶ + ܿߙݏଵ) ࢇ࢟൧
ܰ
ܥ
(1.24)
Case (i)
If P is the midpoint, α1= α2 = α
ࡱ =
ߣ
4ߨߝℎ
ൣ(ߙ ݊݅ݏ − ࢇ)ߙ ݊݅ݏ࢞ + (ܿߙ ݏ + ܿࢇ )ߙ ݏ࢟൧
ܰ
ܥ
ࡱ =
ߣ
4ߨߝℎ
ൣ2 ܿࢇ ߙ ݏ࢟൧
ܰ
ܥ
ࡱ =
ߣ
2ߨߝℎ
ܿࢇ ߙ ݏ࢟
ܰ
ܥ
(1.25)
The direction of electric field intensity is normal to the line charge. The electric field is not
normal to the line charge if the point is not at midpoint.
Case (ii)
As length tends to
α1 = 0 and α2 = 0
from equation (1.24), we get
ࡱ =
ߣ
2ߨߝℎ
ࢇ࢟
ܰ
ܥ
(1.26)
1.3.4) Electrical field due to charged ring:
A circular ring of radius a carries a uniform charge λ C/m and is placed on the xy-plane with
axis the same as the z-axis as shown in figure.
Let dE be the electric field intensity due to a charge dQ. The ring is assumed to be formed by
several point charges. When these vectors are resolved, radial components get cancelled and
normal components get added. Therefore the direction of electric field intensity is normal to
the plane of the ring. The sum of normal components can be written as
∞
11. 1.3.5) Electrical field due to a charged disc:
A disc of radius ‘a’ meters is uniformly charged with a charged density σ c/m2
. It is required
to determine the electric field at ‘P’ which is at a distance h meters from the centre of the disc
as shown in figure 1.7
Fig. 1.7 charged disc
The disc is assumed to be formed by several rings of increasing radius. Consider a ring of
radius x meters. Each ring is assumed to be formed by number of point charges.
Let dE1 be the electric field intensity due to a charge dQ1 and dE2 is the electric field intensity
due to a charge dQ2.
Electric field due to one ring be obtained by adding normal components of dE1, dE2 .......
dEn.
Therefore
dE = (dE1 cosߠ + dE2 cosߠ + ....... + dEn cosߠ ) az
dE = (dE1 + dE2 + ....... + dEn) cosߠ az
dE = (
ௗொభ
ସగఌమ +
ௗொమ
ସగఌమ + ....... +
ௗொ
ସగఌమ) cosߠ az
dE =
ௗொభାௗொమା⋯ାௗொ
ସగఌమ cosߠ az
dE =
ௗொ
ସగఌమ cosߠ az
The total charge of the ring is σ ds which is equal to dQ
12. dE =
σ ୢୱ
ସగఌమ
cosߠ az (1.28)
ds = π[(x + dx)2
– x2
]
ds = π[x2
+ dx2
+ 2xdx – x2
] = 2π x dx (neglecting dx2
term)
Substituting ds in eq. (1.28) , we get
dE =
σ ଶπ୶ୢ୶
ସగఌమ cosߠ az
dE =
σ ୶ୢ୶
ଶఌమ
cosߠ az (1.29)
From above figure we can write
Tanߠ = x/h
x = h tanߠ
(1.30)
dx = h sec2
ߠ dߠ
(1.31)
cosߠ = h/r
r = h/ cosߠ = h secߠ
(1.32)
Substituting eqs. (1.30), (1.31) and (1.32) in (1.29) we get,
dE =
σ ( ୦ ୲ୟ୬θ)(୦ ୱୣୡమθ ୢθ)
ଶఌ୦ ୱୣୡθమ cosߠ az
dE =
σ
ଶఌ
sinߠ dߠ az
On integrating
ࡱ =
ߪ
2ߝ
න ߠ݊݅ݏ
ఈ
݀ߠ ࢇࢠ
ࡱ =
ߪ
2ߝ
ሾ−ܿߠݏሿ
ఈ
ࢇࢠ
ࡱ =
ߪ
2ߝ
(1 − ܿࢇ)ߙݏࢠ (1.33)
From figure 1.7
ܿߙݏ =
ℎ
ඥ(ܽଶ + ℎଶ)
13. ∴ ࡱ =
ߪ
2ߝ
ቆ1 −
ℎ
ඥ(ܽଶ + ℎଶ)
ቇ ࢇࢠ
ܰ
ܥ
(1.34)
For infinite disc, radius ‘a’ tends to infinite and α = 90.
ࡱ =
ߪ
2ߝ
(1 − ܿࢇ)09 ݏࢠ
ࡱ =
ߪ
2ߝ
ࢇࢠ
ܰ
ܥ
ݎ
ܸ
݉
(1.35)
From equation (1.35), it can be seen that electric field due to infinite disc is independent of
distance. Electric field is uniform.
Problems
1. Determine the magnitude of electric field at a point 3 cm away from the mid point of
two charges of 10-8
c and -10-8
c kept 8 cm apart as shown in figure.
Ans. Er = 2 |E| cos ߠ
|E1|=|E2|=E
ܧ =
1
4ߨߝ
ܳ
ܴଶ
ܧ =
1
4ߨ 8.854 × 10ିଵଶ
10ି଼
0.05ଶ
= 36 ݉/ܸ ܭ
Er = 2 36 × 103
cos 53.1 = 57.6 K V
2. A uniform charge distribution infinite in extent lies along z-axis with a charge density
of 20 nc/m. Determine electric field intensity at (6,8,0).
Ans. d = (6-0)i+(8-0)j+0 k = 6i + 8j
d= ඥ(6ଶ + 8ଶ) = 10
ࡱ =
ߣ
2ߨߝ݀
ࢇࢊ
ܰ
ܥ
ࡱ =
20 × 10ିଽ
2ߨ 8.854 × 10ିଵଶ 10
6i + 8j
10
E = 21.57i + 28.76j N/C
3. A straight wire has an uniformly distributed charge of 0.3×10-4
c/m and of length 12
cm. Determine the electric field at a distance of 3 cm below the wire and displaced 3
cm to the right beyond one end as shown in figure.
Ans. tan α1 =
=
.ଷ
.ଵଶା.ଷ
=
.ଷ
.ଵହ
14. α1 = tan-1
(0.03/0.15)= 11.3o
tan β = 0.03/0.03 = 1
β = 45o
,
α2 = 180-45= 135o
ࡱ =
ߣ
4ߨߝℎ
ൣ(ߙ݊݅ݏଶ − ߙ݊݅ݏଵ)ࢇ࢞ + (ܿߙݏଶ + ܿߙݏଵ) ࢇ࢟൧
ࡱ =
9 × 10ଽ
× 0.3 × 10ିସ
0.03
ൣ(sin 135 − sin 11.3)ࢇ࢞ + (cos 135 + cos 11.3) ࢇ࢟൧
ࡱ = 4.59 × 10
ࢇ࢞ + 2.4 × 10
ࢇ࢟
ࡱ = ൫4.59 ࢇ࢞ + 2.4 ࢇ࢟൯ ݉/ݒܭ
4. Find the force on a point charge 50 µc at (0,0,5) m due to a charge 0f 500π µc. It is
uniformly distributed over a circular disc of radius 5 m.
Ans. ߪ =
ொ
గ మ
ߪ =
500π 10ି
ߨ 5ଶ
= 20 µ c/m
ࡱ =
ߪ
2ߝ
(1 − ܿࢇ)ߙݏࢠ
ࡱ =
20 × 10ି
2 × 8.854 × 10ିଵଶ
(1 − ܿࢇ )54 ݏࢠ
ࡱ = 3.3 × 10ହ
ࢇࢠ ܰ/ ܥ
F= E q
F = 3.3 × 10ହ
× 50 × 10ି
= 16.5 ܰ
F = 16.5 az N
5. A charge of 40/3 nc is distributed around a ring of radius 2m, find the electric field at
a point (0,0,5)m from the plane of the ring kept in Z=0 plane.
15. Ans. Q = 40/3 n c
ߣ =
ܳ
2 ߨ ܽ
=
40
3 × 10ିଽ
2 ߨ 2
= 1.06 × 10ିଽ
ܿ
݉
ࡱ =
ߣ ܽ ℎ
2ߝݎଷ
ࢇࢠ
N
C
ࡱ =
1.06 × 10ିଽ
× 2 × 5
2 × 8.854 × 10ିଵଶ × √29
ଷ ࢇࢠ
ࡱ = 3.83 ࢇࢠ ܸ
6. An infinite plane at y=3 m contains a uniform charge distribution of density 10-8
/6π
C/m2
. Determine electric field at all points.
Ans. (i) y>3
E = σ/2ϵ j = 10-8
/(2×6π×8.854×10-12
) j = 29.959 j
Magnitude of electric field at all points is same since the electric field due to an
infinite plane (or) disc is uniform.
For y<3
E = -29.959 j
For y<3, electric field is directed along negative y-axis. Unit vector along negative y-
direction is –j.
7. A straight line of charge of length 12 cm carries a uniformly distributed charge of
0.3×10-6
coulombs per cm length. Determine the magnitude and direction of the
electric field intensity at a point
(i) Located at a distance of a 3 cm above the wire displaced 3 cm to the right of
and beyond one end.
(ii) Located at the distance of 3 cm from one end, in alignment with, but beyond
the wire.
(iii) Located at a distance of 3 cm from one end, on the wire itself.
ans.
(i) tan α1 =
=
.ଷ
.ଵଶା.ଷ
=
.ଷ
.ଵହ
α1 = tan-1
(0.03/0.15)= 11.3o
tan β = 0.03/0.03 = 1
β = 45o
,
α2 = 180-45= 135o
16. ࡱ =
ߣ
4ߨߝℎ
ൣ(ߙ݊݅ݏଶ − ߙ݊݅ݏଵ)ࢇ࢞ + (ܿߙݏଶ + ܿߙݏଵ) ࢇ࢟൧
ࡱ =
9 × 10ଽ
× 0.3 × 10ିସ
0.03
ൣ(sin 135 − sin 11.3)ࢇ࢞ + (cos 135 + cos 11.3) ࢇ࢟൧
ࡱ = 4.59 × 10
ࢇ࢞ + 2.4 × 10
ࢇ࢟
ࡱ = ൫4.59 ࢇ࢞ + 2.4 ࢇ࢟൯ ݉/ݒܭ
(ii) consider an element dx of charge λdx at a distance x from A. The field at P due to
the positive charge element λdx is directed to the right.
ࢊࡱ =
ߣ݀ݔ
4ߨߝ(ܮ + ℎ − )ݔଶ
ࢇ࢞
ࡱ =
ߣ
4ߨߝ
න
݀ݔ
(ܮ + ℎ − )ݔଶ
ࡸ
ࢇ࢞
ࡱ =
ߣ
4ߨߝ
1
ℎ
−
1
ܮ + ℎ
൨ ࢇ࢞
ࡱ = 9 × 10ଽ
× 0.3 × 10ିସ
1
0.03
−
1
0.15
൨ ࢇ࢞
ࡱ = 7.2 × 10
ࢇ࢞ ࢜/
(iii) Now L=AC= 6 cm = 0.06 cm
h = 0.03 m
ࡱ =
ߣ
4ߨߝ
1
ℎ
−
1
ܮ + ℎ
൨ ࢇ࢞
ࡱ = 9 × 10ଽ
× 0.3 × 10ିସ
1
0.03
−
1
0.09
൨ ࢇ࢞
17. E = 60 ax K v/cm
1.4) Work Done:
If a point charge ‘Q’ is kept in an electric field it experience a force F in the direction of
electric field. Fa is the applied force in a direction opposite to that of F.
Let dw be the work done in moving this charge Q by a distance dl m. Total work done in
moving the point charge from ‘a’ to ‘b’ can be obtained by integration.
ܹ = න ݀ ݓ
ܹ = න ࡲࢇ ∙ ࢊ
ܹ = − න ࡲ ∙ ࢊ
(1.36)
E = F/Q
F = E Q
ܹ = − න ࡱ ܳ ∙ ࢊ
ܹ = −ܳ න ࡱ ∙ ࢊ
(1.37)
E = Ex ax + Ey ay + Ez az
dl = dx ax + dyay + dz az
E ∙ dl = EX dx + Ey dy + Ez dz
∴ ܹ = −ܳ න (E୶ dx + E୷ dy + E dz)
(௫మ,௬మ,௭మ)
(௫భ,௬భ,௭భ)
(1.38)
Problems
18. 1. Find the work involved in moving a charge of 1 c from (6,8,-10) to (3,4,-5) in the
field E = -x i +y j –3 k.
Ans.
ܹ = −ܳ න (E୶ dx + E୷ dy + E dz)
(ଷ.ସ.ିହ)
(,଼,ିଵ)
ܹ = −1 ቈන −ݔ݀ ݔ +
ଷ
න ݕ݀ ݕ
ସ
଼
− න ݖ݀ ݖ
ିହ
ିଵ
ܹ = − ቆ
−ݔଶ
2
ቇ
ଷ
+ ቆ
ݕଶ
2
ቇ
଼
ସ
− (3)ݖିଵ
ିହ
൩
ܹ = −
−9
2
+
36
2
+
16
2
−
64
2
− −15 + 30൨
W= 25.5 J
2. Find the work done in moving a point charge of -20 µc from the origin to (4,0,0) in
the field E = (x/2 + 2y) i + 2x j
Ans.
ܹ = −ܳ න (E୶ dx + E୷ dy + E dz)
(௫మ,௬మ,௭మ)
(௫భ,௬భ,௭భ)
dl = d xi
ܹ = −ܳ න (E୶ dx )
(ସ,,)
(,,)
ܹ = 20 × 10ି
න ((x/2 + 2y) dx )
(ସ,,)
(,,)
ܹ = 20 × 10ି
ቆ
ݔଶ
4
ቇ
ସ
W = 80×10-6
J
1.5) Absolute Potential:
Absolute potential is defined as the work done in moving a unit positive charge from infinite
to the point against the electric field.
19. A point charge Q is kept at an origin as shown in figure. It is required to find the potential at
‘b’ which is at distance ‘r’ m from the reference.
Consider a point R1 at a distance ‘x’ m. The small work done to move the charge from R1 to
P1 is dw. The electrical field due to a point charge Q at a distance ‘x’ mis
ࡱ = ܳ/(4ߨߝݔଶ
)
Work done = E• dx
Total work done can be obtained by integration
Work done (W) = − Q ۳ ∙ dܔ
ୠ
ୟ
V= − 1 ۳ ∙ dܔ
ୠ
ୟ
V = − න ۳ ∙ dܔ
ୠ
ୟ
E = Ex ax , dl = dx ax
V = − න
ܳ
4ߨߝݔଶ
݀ݔ
ୠ
ୟ
ܸ =
ܳ
4ߨߝݎ
(1.39)
1.6) Potential Difference: Vab
Potential difference Vab is defined as the work done in moving a unit positive charge from ‘b’
to ‘a’.
20. Consider a point charge Q kept at the origin of a spherical co-ordinate system. The
field is always in the direction of ar. No field in the direction of ߠ and ϕ. The points ‘a’ and
’b’ are at distance ra and rb respectively as shown in figure.
Vୟୠ = − න ۳ ∙ dܔ
ୟ
ୠ
E = Er ar , dl = dr ar
Vୟୠ = − න E୰d୰
ୟ
ୠ
Vୟୠ = − න
ܳ
4ߨߝݎଶ
d୰
୰
୰ౘ
Vୟୠ =
ܳ
4ߨߝ
1
ݎ
−
1
ݎ
൨
Vୟୠ =
ܳ
4ߨߝ
1
ݎ
−
ܳ
4ߨߝ
1
ݎ
Vୟୠ = Vୟ − ܸ (1.40)
1.6.1) Potential difference due to line charge:
The wire is uniformly charged with λ C/m. We have to find the potential difference Vab due
to this line charge. Consider a point P at a distance P from the line charge.
21. Fig. Line charge
E = Eρ aρ
dl = dρ aρ
E•dl = Eρ aρ • dρ aρ = Eρ dρ
Potential difference Vab is the work done in moving a unit +ve charge from ‘b’ to ‘a’.
Vୟୠ = − න ۳ ∙ dܔ
ୟ
ୠ
Vୟୠ = − න Eρ dρ
ୟ
ୠ
Vୟୠ = − න
ߣ
2ߨߝߩ
dρ
ρ
ρౘ
Vୟୠ =
ߣ
2ߨߝ
ln
ߩ
ߩ
(1.41)
1.6.2) Potential due to charged ring:
A thin wire is bent in the form of a circular ring as shown in figure. It is uniformly charged
with a charge density λ C/m. It is required to determine the potential at height ‘h’ m from the
centre of the ring. The ring is assumed to be formed by several point charges.
Fig. Charged ring
Let dv be the potential due to the line charge element of length dl containing a charge
dQ.
݀ݒ =
݀ܳ
4ߨߝݎ
݀ݒ =
ߣ݈݀
4ߨߝݎ
ܸ = න
ߣ݈݀
4ߨߝݎ
22. ܸ =
ߣ
4ߨߝݎ
න ݈݀
ܸ =
ߣ
4ߨߝݎ
2ߨܽ
ܸ =
ߣܽ
2ߝݎ
ܸ =
ߣܽ
2ߝ√ܽଶ + ܾଶ
)24.1( ݏݐ݈ݒ
1.6.3) Potential due to a charged disc:
let dv be the potential due to one ring. Each ring is assumed to be having several point
charges dQ1, dQ2, ............ dQn. Potential due to the ring is the sum of potential.
Fig. Charge disc
݀ݒ =
݀ܳଵ
4ߨߝݎ
+
݀ܳଶ
4ߨߝݎ
+ … … … … … . +
݀ܳ
4ߨߝݎ
݀ݒ =
݀ܳଵ + ݀ܳଶ + … … . . + ݀ܳ
4ߨߝݎ
݀ݒ =
݀ܳ
4ߨߝݎ
݀ݒ =
ߪ ݀ݏ
4ߨߝݎ
=
ߪ 2ߨݔ݀ ݔ
4ߨߝݎ
=
ߪ ݔ݀ ݔ
2ߝݎ
Potential due to entire disc can be obtained by integration
ܸ = න ݀ݒ = න
ߪ ݔ݀ ݔ
2ߝݎ
=
ߪ
2ߝ
න
ݔ
√ݔଶ + ℎଶ
݀ ݔ
Let ݔଶ
+ ℎଶ
= ݐଶ
2x dx = 2t dt
Therefore we have
23. ܸ =
ߪ
2ߝ
න
ݐ݀ ݐ
ݐ
√మାమ
ܸ =
ߪ
2ߝ
න ݀ݐ
√మାమ
ܸ =
ߪ
2ߝ
ቀඥܽଶ + ℎଶ − ℎቁ )34.1( ݏݐ݈ݒ
At the centre of the disc , h=0;
ܸ =
ߪܽ
2ߝ
)44.1( ݏݐ݈ݒ
Problems
1. Find the work done in moving a poimt charge Q= 5 µ c from origin to (2m, π/4 , π/2),
spherical co-ordinates in the field E = 5 e-r/4
ar + (10/(r sin θ)) aϕ v/m.
Ans.
dl = dr ar + rdθ aθ + r sinθ dϕ aϕ
E• dl = 5 e-r/4
dr + 10 dϕ
W = − Q න ۳ ∙ dܔ
ୠ
ୟ
W = − 5 × 10ି
ቈන 5 eି୰/ସ
dr
ଶ
න 10 dϕ
π/ଶ
W= -117.9 µ J.
2. 5 equal, point charges of 20 nc are located at x=2,3,4,5 and 6 cm. Determine the
potential at the origin.
Ans.
By superposition principal
V= V1 + V2 + V3 + V4 + V5
ܸ =
ܳ
4ߨߝݎଵ
+
ܳ
4ߨߝݎଶ
+
ܳ
4ߨߝݎଷ
+
ܳ
4ߨߝݎସ
+
ܳ
4ߨߝݎହ
24. ܸ =
20 × 10ି
4ߨߝ
൬
1
2
+
1
3
+
1
4
+
1
5
+
1
6
൰ = 261 ݏݐ݈ݒ
3. A line charge of 0.5 nc/m lies along z-axis. Find Vab if a is (2,0,0) , and b is (4,0,0).
Ans.
Vୟୠ =
ߣ
2ߨߝ
ln
ߩ
ߩ
Vୟୠ =
0.5 × 10ିଽ
2ߨ × 8.854 × 10ିଵଶ
ln
4
2
Vab = 6.24 V.
4. A point charge of 0.4 nc is located at (2,3,3) in Cartesian coordinates. Find Vab if a is
(2,2,3) and b is (-2,3,3).
Ans.
Vୟୠ = Vୟ − ܸ
Vୟୠ =
ܳ
4ߨߝ
1
ݎ
−
ܳ
4ߨߝ
1
ݎ
ra = (2-2) i + (2-3) j + (3-3) k = -j
ra = 1
rb = (-2-2) i + (3-3) j + (3-3) k = -4 i
rb = 4
Vୟୠ =
0.4 × 10ିଽ
4ߨ × 8.854 × 10ିଵଶ
1
1
−
1
4
൨ = 2.7 ݒ
5. A total charge of 40/3 nc is distributed around a ring of radius 2m. Find the potential
at a point on the z-axis 5m from the plane of the ring kept at z=0. What would be the
potential if the entire charge is concentrated at the centre.
ans.
ܸ =
ߣܽ
2ߝ√ܽଶ + ܾଶ
ݏݐ݈ݒ
ߣ =
ܳ
݈
=
40
3
× 10ିଽ
2ߨ 2
= 1.06 × 10ିଽ
ܥ
݉
25. ܸ =
1.06 × 10ିଽ
× 2
2 × 8.854 × 10ିଵଶ × √2ଶ + 5ଶ
= 22.2 ݒ
6. Determine the potential at (0,0,5)m due to a total charge of 10-8
c distributed uniformly
along a disc of radius 5m lying in the Z=0 plane and centered at the orign.
Ans.
ܸ =
ߪ
2ߝ
ቀඥܽଶ + ℎଶ − ℎቁ ݏݐ݈ݒ
ߪ =
ܳ
ܽܽ݁ݎ
=
10ି଼
ߨ × 5ଶ
ܸ =
10ି଼
ߨ × 5ଶ
2ߝ
ቀඥ5ଶ + 5ଶ − 5ቁ
V= 14.89 v
1.7) Relation between V and E:
Consider a point charge Q at the origin as shown in figure. Electric field due to this charge at
the point ‘P’ is
ࡱ =
ܳ
4ߨߝݎଶ
ࢇ࢘
Consider
સ ൬
1
ݎ
൰ = ൬
߲
߲ݎ
ࢇ࢘ + … … … ൰ ൬
1
ݎ
൰
સ ൬
1
ݎ
൰ = − ൬
1
ݎଶ
൰ ࢇ࢘
Using above expression in E , we get
ࡱ =
ܳ
4ߨߝ
− સ ൬
1
ݎ
൰
ࡱ = −સ
ܳ
4ߨߝ
൬
1
ݎ
൰
Therefore,
ࡱ = −સ ܸ
26. Problem
1. Given V= (x-2)2
× (y+2)2
×(z-1)3
. Find electric field at the origin.
Ans.
V= (x-2)2
× (y+2)2
×(z-1)3
ࡱ = −સ ܸ
ࡱ = − ൬
∂v
∂x
i +
∂v
∂y
j +
∂v
∂z
k൰
ࡱ = −2(x − 2)× (y+2)2
×(z-1)3
i-(x-2)2
× 2(y+2) ×(z-1)3
j - (x-2)2
× (y+2)2
×3(z-1)2
k
Electric field at the origin
ࡱ = −2(0 − 2) (0+2)2
×(0-1)3
i -(0-2)2
× 2(0+2) ×(0-1)3
j - (0-2)2
× (0+2)2
×3(0-1)2
k
E = -(16i+16j+48k) v/m.
1.8) Poisson’s and Laplace’s Equations:
From the Gauss law we know that
.ܦ ݀ݏ = Q (1)
A body containing a charge density ρ uniformly distributed over the body. Then charge of that body
is given by
Q = ૉ ݀ݒ (2)
.ܦ ݀ݏ = ૉ ݀ݒ (3)
This is integral form of Gauss law.
As per the divergence theorem
.ܦ ݀ ݏ = સ. ۲ ݀ݒ (4)
સ. ۲ = ૉ (5)
This is known as point form or vector form or polar form. This is also known as Maxwell’s first
equation.
D = ƐE (6)
સ. ƐE = ρ
સ. E = ࣋/Ɛ (7)
We know that E is negative potential medium
27. E = -સ܄ (8)
From equations 7 and 8
સ. (– સ)܄ = ࣋/Ɛ
સ2
V = - ࣋/Ɛ (9)
Which is known as Poisson’s equation in static electric field.
Consider a charge free region (insulator) the value of ρ = 0, since there is no free charges in dielectrics
or insulators.
સ2
V = 0
This is known as Laplace’s equation.