Electric Fields

Electrostatics. When certain types of object are rubbed together electrons are transferred from one surface to another, causing the objects to become charged. The object which gains electrons becomes negatively charged and the object that looses electrons gains a positive charge. If two balloons where charged by rubbing them on a jumper, then hung side by side from a fine thread what would you expect to happen to the Balloons? Why?

We already know that like charges repel and opposite charges attract. Coulomb conducted experiments into the forces of attraction and repulsion between charged objects in the 1780’s. The resulting law is called Coulombs law and describes the behaviour of charged particles in a similar way to Newton’s laws explaining the force of gravity between objects.

However because electrostatic forces can be either attractive or repulsive there are a few differences. The constants have also changed.

Coulomb’s law sates: The magnitude of a force (F) between two electrically charged bodies, which are small compared to their separation (r), is inversely proportional to r2, and proportional to the product of their charges (Q1 and Q2)

Q 1 Q 2 r Notice must be paid to the charges of the bodies, the charges may not be the same magnitude or have the same polarity (+ve or –ve)

When we add in our constants this becomes: Where: F = force (N) Q1 and Q2 = charges (C) r = Separation (m)  = Permittivity in farads per metre (Fm-1) The permittivity of free space (vacuum) is given by  0 and its value is determined by experiment.

 0 is linked to the speed of light in a vacuum (2.998 x 10 8 ) and the permeability of free space (  0 = 4  x10 -7 ) by the equation: Constants:  0 = 8.854 x 10-12 or = 8.988 x 10 9 m/F

Compare Coulomb’s law to Newton's law of gravitation, what similarities and differences do you notice?

Electric field strength E An electric field exists in a region if electrical forces are exerted on charged bodies or particles in that region. The direction of an electric field at a point is the direction in which a small positive charge would move under the influence of the field, if placed at that point.

Field strength due to a point charge. A point charge gives off a radial electric field (similar to planets causing radial gravitational fields). We can calculate the strength of the force experienced by Q’ using: 1 r Point charge (+Q) E Test charge (+Q’)

The electric field intensity is the force per unit charge, so the electric field strength at Q’ is given by: If we substitute 1 into 2 this give us: 2 Radial field

Radial fields (draw in the radial field lines) Q+ Q-

Electric field strength in a uniform field. If we consider two parallel conducting plates, A and B, a certain distance apart (d), There is a potential difference (V) between the plates. The field between these plates will be uniform. Separation = d P.d. = V A B

The field is uniform between the two plates, with all field lines parallel (except at the edges). There is a potential gradient between A and B is the same at any point, except at the edges, the potential at any point is therefore equal to the average potential , the size of which is V/d. Therefore the electric field strength E is given by: Uniform field

Questions. 1. Hydrogen’s single electron orbits about 0.5 Å (5 x 10-11 m) from the centre of the nucleus. Work out the Force acting between the nucleus and the electron. Work out the field strength at the distance r. 2. Chlorine’s outer shell is about 2 x 10-10 m from the centre of its nucleus. Work out the force of attraction between the nucleus and an electron in the inner and outer shells. What is the difference in electric field strength between the 2 orbits?

3. The dome of a van de Graff generator has a charge of 20 C, the dome is 20 cm across. Calculate the field strength, 1mm above the dome Calculate the field strength at a distance of 30 cm If a sphere with a negative charge of 3C was placed 15 cm from the dome, how much force would it experience? 4. An Scanning Electron Microscope accelerates the electrons using plates with holes in them (assume the plates are parallel and creating a uniform field), the accelerating voltage is 8.5 kV the plates are 12 cm apart. Calculate the electric field strength. If the accelerating voltage is dropped to 5kV what is the field strength.

Electric potential In order to move a charge from one point to another, work may have to bed done on the charge. For this to happen there must be some difference between the properties at the two points. This property is called electric potential, the concept of which is similar to gravitational potential. The electric potential is defined as: The potential at a point in an electric field is defined as being numerically equal to the work done in bringing a positive charge from infinity to the point. Electric potential is measured in volts (V).

The formula for calculating the electric potential is very similar to the equations for determining the force experience by charges and the electric field strength due to a charge. (Just as the equation for gravitational potential is closely related to the equations for Newton’s laws of gravitation and g in a radial field) Where: V = Electric potential. (V)  0 = Permittivity of free space (F/m) r = separation (m) Q = Charge (C)

Field lines and Equipotential An electric field is a region around a charge where another charge will experience a force. Unlike gravitational fields electric fields can be attractive or repulsive. Draw field lines on the diagrams below. The lines of equipotential link areas with exactly the same electric potential and are always at right angles to the Field lines. (gravitational equipotentials follow the same principle) + - +Q -Q

Motion of charged particles in an electric field. The idea of electric potential can be used to explain the acceleration of charged particles. From the definition of a volt (V = W/Q), we can say that the kinetic energy gained by a particle carrying a charge q when accelerated through a P.D. of V volts is: Where m is the mass of the particle and v is its subsequent speed.

We can use this equation to work out the speed of an electron as it leaves an electron gun, as used in CRT and television tubes. Calculate the speed of an electron if it is accelerated by PD of 5kV.

In CRTs in order to form the images the electrons are deflected to different parts of the screen using electric fields. As we know already, charged particles are deflected by electric fields and the force experienced by a particle is determined by its charge q and the field strength E . and Where q is the minor or moving charge (Q 2)

In a uniform field such as those created by a set of deflection plates such as those in a CRT or TV the field strength is equal to E = v/d so: Where V is the PD between the plates, d is their separation and q is the charge of the particle.

Calculate the force deflecting the particles (  ) if the PD is 2kV and the plates are 6cm apart. Beam of +ve Particles + -

Electron deflection tube: We will use this piece of apparatus to investigate the defecting force of electric fields on a stream of electrons, we will use blue electrons for this experiment.

Points to note: The force is always parallel to the field lines. The path of the particles is parabolic within the electric field and straight at all other times. The size of the force is independent of speed. +Ve and –Ve particles are deflected in opposite directions. Gravity is ignored. Not on syllabus but worth noting: Deflection of particles in a magnetic field follows Fleming’s left hand rule. Particles in a magnetic field follow a circular path.

Questions 1 A point charge of +4nC is brought from infinity to a point 30mm away from a charge of +8nC. How much work is done? How much more work must be done to bring the charges 10mm closer together? 2 In a CRT an electron is accelerated though a PD of 2.5kV what is its velocity when it exits the gun? If the plates are 5cm long how long does it take for the electron to pass between them? If the field strength between the plates is 10kV m -1 how big is the deflecting force? plates 5cm separation

Question 3. Calculate the gravitational and electrical force on 2 electrons 1.00 x 10 -10 m apart in a vacuum. What do you notice about the forces? Calculate the ratio between the forces. Other magnitude state one difference between these forces.

Can be used to describe the field – direction is the direction of the force on a positive charge Can be used to describe the field – direction is the direction of the force on a mass Field lines Infinite, decreases with distance but theoretically never reaches zero. Infinite, decreases with distance but theoretically never reaches zero. Range of field Contact not needed, acts at distance Contact not needed, acts at distance For the force to act Electrical Gravitational

Can be used to describe work done in moving charges: (inverse law) Can be used to describe work done in moving masses: (inverse law) Potential Given by Coulomb’s law: Given by Newton’s law of universal gravitation: Force between 2 objects. Force on a unit charge: Force on a unit mass: Field strength

Kinetic energy Field strength is (-ve of) potential gradient. Field strength is (-ve of) potential gradient. Relationship between potential and field strength. Electrical PE: GPE: Potential energy

Differences Comparative sizes Shielding Effect Produced by and acts on charges Produced by and acts on masses Origin

Electric Fields

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