1. ELECTROSTATICS - I – Electrostatic Force1. Frictional Electricity2. Properties of Electric Charges3. Coulomb’s Law4. Coulomb’s Law in Vector Form5. Units of Charge6. Relative Permittivity or Dielectric Constant7. Continuous Charge Distribution i) Linear Charge Density ii) Surface Charge Density iii) Volume Charge Density
2. Frictional Electricity:Frictional electricity is the electricity produced by rubbing two suitable bodiesand transfer of electrons from one body to other. + ++ - .- - - - - + ++ + + - - Ebonite ++ - - ++ Glass + ++ + Flannel Silk + ++ + ++Electrons in glass are loosely bound in it than the electrons in silk. So, whenglass and silk are rubbed together, the comparatively loosely bound electronsfrom glass get transferred to silk.As a result, glass becomes positively charged and silk becomes negativelycharged.Electrons in fur are loosely bound in it than the electrons in ebonite. So, whenebonite and fur are rubbed together, the comparatively loosely bound electronsfrom fur get transferred to ebonite.As a result, ebonite becomes negatively charged and fur becomes positivelycharged.
3. It is very important to note that the electrification of the body (whetherpositive or negative) is due to transfer of electrons from one body to another.i.e. If the electrons are transferred from a body, then the deficiency ofelectrons makes the body positive.If the electrons are gained by a body, then the excess of electrons makes thebody negative.If the two bodies from the following list are rubbed, then the body appearingearly in the list is positively charges whereas the latter is negatively charged.Fur, Glass, Silk, Human body, Cotton, Wood, Sealing wax, Amber, Resin,Sulphur, Rubber, Ebonite. Column I (+ve Charge) Column II (-ve Charge) Glass Silk Wool, Flannel Amber, Ebonite, Rubber, Plastic Ebonite Polythene Dry hair Comb
4. Properties of Charges:1. There exists only two types of charges, namely positive and negative.2. Like charges repel and unlike charges attract each other.3. Charge is a scalar quantity.4. Charge is additive in nature. eg. +2 C + 5 C – 3 C = +4 C5. Charge is quantized. i.e. Electric charge exists in discrete packets rather than in continuous amount. It can be expressed in integral multiples fundamental electronic charge (e = 1.6 x 10-19 C) q = ± ne where n = 1, 2, 3, …………6. Charge is conserved. i.e. The algebraic sum of positive and negative charges in an isolated system remains constant. eg. When a glass rod is rubbed with silk, negative charge appears on the silk and an equal amount of positive charge appear on the glass rod. The net charge on the glass-silk system remains zero before and after rubbing. It does not change with velocity also.
5. Note: Recently, the existence of quarks of charge ⅓ e and ⅔ e has beenpostulated. If the quarks are detected in any experiment with concretepractical evidence, then the minimum value of ‘quantum of charge’ will beeither ⅓ e or ⅔ e. However, the law of quantization will hold good.Coulomb’s Law – Force between two point electric charges:The electrostatic force of interaction (attraction or repulsion) between two pointelectric charges is directly proportional to the product of the charges, inverselyproportional to the square of the distance between them and acts along the linejoining the two charges.Strictly speaking, Coulomb’s law applies to stationary point charges. q1 q2 F α q1 q2 F α 1 / r2 r q1 q2 q1 q2 where k is a positive constant of or Fα or F=k proportionality called r2 r2 electrostatic force constant or Coulomb constant. 1 In vacuum, k = where ε0 is the permittivity of free space 4πε0
6. 1 In medium, k = where ε is the absolute electric permittivity of 4πε the dielectric mediumThe dielectric constant or relative permittivity or specific inductive capacity ordielectric coefficient is given by ε K = εr = ε0 1 q1 q2 In vacuum, F = 4πε0 r2 1 q1 q2 In medium, F = 4πε0εr r2 ε0 = 8.8542 x 10-12 C2 N-1 m-2 1 1 = 8.9875 x 109 N m2 C-2 or = 9 x 109 N m2 C-2 4πε0 4πε0
7. Coulomb’s Law in Vector Form:In vacuum, for q1 q2 > 0, + q1 r12 + q2 F12 r F21 1 q1 q2 F12 = r21 4πε0 r2 q1q2 > 0 1 q1 q2 F21 = r12 - q1 - q2 4πε0 r2 r12 F12 r F21 q1q2 > 0In vacuum, for q1 q2 < 0, + q1 r12 - q2 1 q1 q2 1 q1 q2F12 = r12 & F = r21 F12 F21 21 4πε0 r2 4πε0 r2 r F12 = - F21 (in all the cases) q1q2 < 0
8. 1 q1 q2 1 q1 q2 F12 = r12 & F21 = r21 4πε0 r3 4πε0 r3Note: The cube term of the distance is simply because of vector form. Otherwise the law is ‘Inverse Square Law’ only.Units of Charge:In SI system, the unit of charge is coulomb (C).One coulomb of charge is that charge which when placed at rest in vacuum at adistance of one metre from an equal and similar stationary charge repels it andis repelled by it with a force of 9 x 109 newton.In cgs electrostatic system, the unit of charge is ‘statcoulomb’ or ‘esu of charge’.In cgs electrostatic system, k = 1 / K where K is ‘dielectric constant’.For vacuum, K = 1. q1 q2 F= r2If q1 = q2 = q (say), r = 1 cm and F = 1 dyne, then q = ± 1 statcoulomb.In cgs electromagnetic system, the unit of charge is ‘abcoulomb’ or ‘emu ofcharge’.
9. 1 emu of charge = c esu of charge 1 emu of charge = 3 x 1010 esu of charge 1 coulomb of charge = 3 x 109 statcoulomb 1 abcoulomb = 10 coulombRelative Permittivity or Dielectric Constant or Specific InductiveCapacity or Dielectric Coefficient:The dielectric constant or relative permittivity or specific inductive capacity ordielectric coefficient is given by the ratio of the absolute permittivity of themedium to the permittivity of free space. ε K = εr = ε0The dielectric constant or relative permittivity or specific inductive capacity ordielectric coefficient can also be defined as the ratio of the electrostatic forcebetween two charges separated by a certain distance in vacuum to theelectrostatic force between the same two charges separated by the samedistance in that medium. Fv K = εr = Dielectric constant has no unit. Fm
10. Continuous Charge Distribution:Any charge which covers a space with dimensions much less than its distanceaway from an observation point can be considered a point charge.A system of closely spaced charges is said to form a continuous chargedistribution.It is useful to consider the density of a charge distribution as we do for densityof solid, liquid, gas, etc.(i) Line or Linear Charge Density ( λ ):If the charge is distributed over a straight line or over the circumference of acircle or over the edge of a cuboid, etc, then the distribution is called ‘linearcharge distribution’.Linear charge density is the charge per unit length. Its SI unit is C / m. q dq λ= or λ= dq l dl ++++++++++++ dl Total charge on line l, q = ∫ λ dl l
11. (ii) Surface Charge Density ( σ ):If the charge is distributed over a surface area, then the distribution is called‘surface charge distribution’.Surface charge density is the charge per unit area. Its SI unit is C / m2. q dq dq ++++++++++++ σ= or σ= S dS ++++++++++++ ++++++++++++ Total charge on surface S, q = ∫ σ dS ++++++++++++ dS S(iii) Volume Charge Density ( ρ ):If the charge is distributed over a volume, then the distribution is called‘volume charge distribution’.Volume charge density is the charge per unit volume. Its SI unit is C / m3. q dq ρ= or ρ= ז dז dq Total charge on volume ,ז q = ∫ ρ dז dז ז
12. ELECTROSTATICS - II : Electric Field 1. Electric Field 2. Electric Field Intensity or Electric Field Strength 3. Electric Field Intensity due to a Point Charge 4. Superposition Principle 5. Electric Lines of Force i) Due to a Point Charge ii) Due to a Dipole iii) Due to a Equal and Like Charges iv) Due to a Uniform Field 6. Properties of Electric Lines of Force 7. Electric Dipole 8. Electric Field Intensity due to an Electric Dipole 9. Torque on an Electric Dipole 10. Work Done on an Electric Dipole
13. Electric Field:Electric field is a region of space around a charge or a system of chargeswithin which other charged particles experience electrostatic forces.Theoretically, electric field extends upto infinity but practically it is limited to acertain distance.Electric Field Strength or Electric Field Intensity or Electric Field:Electric field strength at a point in an electric field is the electrostatic force perunit positive charge acting on a vanishingly small positive test charge placedat that point. +q + q0 -q + q0 F F q – Source charge, q0 – Test charge, F – Force & E - Field 1 q Lt F F or E= r E= or E= r2 ∆q → 0 ∆q q0 4πε0The test charge is considered to be vanishingly small because its presenceshould not alter the configuration of the charge(s) and thus the electric fieldwhich is intended to be measured.
14. Note:1. Since q0 is taken positive, the direction of electric field ( E ) is along the direction of electrostatic force ( F ).2. Electrostatic force on a negatively charged particle will be opposite to the direction of electric field.3. Electric field is a vector quantity whose magnitude and direction are uniquely determined at every point in the field.4. SI unit of electric field is newton / coulomb ( N C-1 ).
15. Electric Field due to a Point Charge: YForce exerted on q0 by q is 1 q q0 F F= r + q0 4πε0 r2 r P (x,y,z) 1 q q0 or F= r 4πε0 r3 +q O X FElectric field strength is E= q0 Z 1 q E (r) = r 4πε0 r3 E 1 q or E (r) = r 4πε0 r2The electric field due to a point charge hasspherical symmetry.If q > 0, then the field is radially outwards. 0 r2If q < 0, then the field is radially inwards.
16. Electric field in terms of co-ordinates is given by 1 q E (r) = ( xi + y j + z k ) 4πε0 ( x2 + y2 + z2 ) 3/2 F14Superposition Principle:The electrostatic force experienced by a - q5charge due to other charges is the vector + q1 + q2sum of electrostatic forces due to these F15other charges as if they are existingindividually. F12 F13 F1 = F12 + F13 + F14 + F15 + q4 - q3 N 1 ra - rb Fa (ra) = ∑ qa qb F12 4πε0 F1 b=1 │ ra - rb │3 b≠a F15In the present example, a = 1 and b = 2 to 5. F13If the force is to be found on 2nd charge, F14then a = 2 and b = 1 and 3 to 5.
17. Note:The interactions must be on the charge which is to be studied due to othercharges.The charge on which the influence due to other charges is to be found isassumed to be floating charge and others are rigidly fixed.For eg. 1st charge (floating) is repelled away by q2 and q4 and attracted towardsq3 and q5.The interactions between the other charges (among themselves) must beignored. i.e. F23, F24, F25, F34, F35 and F45 are ignored.Superposition principle holds good for electric field also.Electric Lines of Force:An electric line of force is an imaginary straight or curved path along which aunit positive charge is supposed to move when free to do so in an electricfield.Electric lines of force do not physically exist but they represent real situations. E E Electric Lines of Force
18. 1. Electric Lines of Force due to a Point Charge: a) Representation of electric field in terms of field vectors: The size of the arrow represents the strength of electric field. q>0 q<0 b) Representation of electric field in terms of field lines (Easy way of drawing)
19. 2. Electric Lines of Force due to a 3. Electric Lines of Force due to a pair of Equal and Unlike Charges: pair of Equal and Like Charges: (Dipole) +q P +q E .N +q -q Electric lines of force contract Electric lines of force exert lateral lengthwise to represent attraction (sideways) pressure to represent between two unlike charges. repulsion between two like charges.
20. 4. Electric Lines of Force due to a Uniform Field: E + - + - Properties of Electric Lines of Force + - or Field Lines: + +1 C -1. The electric lines of force are imaginary lines.2. A unit positive charge placed in the electric field tends to follow a path along the field line if it is free to do so.3. The electric lines of force emanate from a positive charge and terminate on a negative charge.4. The tangent to an electric field line at any point . E P gives the direction of the electric field at that point.5. Two electric lines of force can never cross each other. If they do, then at the point of intersection, there will be two tangents. It means there are two values of the electric field E1 at that point, which is not possible. LEE SIB Further, electric field being a vector quantity, OS E2 there can be only one resultant field at the TP given point, represented by one tangent at the NO given point for the given line of force.
21. 6. Electric lines of force are closer (crowded) where the electric field is stronger and the lines spread out where the electric field is Q weaker. q7. Electric lines of force are perpendicular to the surface of a positively or negatively charged Q > q body.8. Electric lines of force contract lengthwise to represent attraction between two unlike charges.9. Electric lines of force exert lateral (sideways) pressure to represent repulsion between two like charges.
22. 10. The number of lines per unit cross – sectional area perpendicular to the field lines (i.e. density of lines of force) is directly proportional to the magnitude of the intensity of electric field in that region. ∆N α E ∆A11. Electric lines of force do not pass through a conductor. Hence, the interior of the conductor is free from the influence of the electric field. E E + - + - - + Solid or hollow + - - + + conductor - (Electrostatic Shielding) - + + - No Field - + - + + -12. Electric lines of force can pass through an insulator.
23. Electric Dipole:Electric dipole is a pair of equal and opposite charges separated by a verysmall distance.The electric field produced by a dipole is known as dipole field.Electric dipole moment is a vector quantity used to measure the strength of anelectric dipole. p p = (q x 2l) l -q +q 2lThe magnitude of electric dipole moment is the product of magnitude of eithercharge and the distance between the two charges.The direction is from negative to positive charge.The SI unit of ‘p’ is ‘coulomb metre (C m)’.Note:An ideal dipole is the dipole in which the charge becomes larger and largerand the separation becomes smaller and smaller.
24. Electric Field Intensity due to an Electric Dipole:i) At a point on the axial line: EP = EB - EAResultant electric field intensityat the point P is A B EA EB EP = EA + EB -q O +q P pThe vectors EA and EB arecollinear and opposite. l l x │EP │ = │EB│ - │EA│ 1 2px 1 q │EP │ = EA = i 4πε0 (x2 – l2)2 4πε0 (x + l)2 q 1 2px 1 EP = i EB = i 4πε0 4πε0 (x - l)2 (x2 – l2)2 1 q q 2p │EP │ = 4πε0 [ (x - l) 2 - (x + l)2 ] If l << x, then EP ≈ 4πε0 x3 The direction of electric field intensity 1 2 (q . 2l) x at a point on the axial line due to a │EP │ = dipole is always along the direction of 4πε0 (x2 – l2)2 the dipole moment.
25. ii) At a point on the equatorial line:Resultant electric field intensity EBat the point Q is EB EB sin θ EQ = EA + EB θ EQ Q EB cos θ θThe vectors EA and EB are θ Q EQacting at an angle 2θ. EA EA cos θ θ q y EA sin θ 1 EA EA = i 4πε0 ( x2 + l2 ) A θ θ B -q O +q 1 q p EB = i 4πε0 ( x2 + l2 ) l lThe vectors EA sin θ and EB sin θ qare opposite to each other and 2 l EQ =hence cancel out. 4πε0 ( x2 + l2 ) ( x2 + l2 )½The vectors EA cos θ and EB cos θ 1 q . 2lare acting along the same direction EQ = 4πε0 ( x2 + l2 )3/2and hence add up. 1 p EQ = EA cos θ + EB cos θ EQ = 4πε0 ( x2 + l2 )3/2
26. 1 p EQ = (- i ) 4πε0 ( x2 + l2 )3/2If l << y, then p EQ ≈ 4πε0 y3The direction of electric field intensity at a point on the equatorial line due to adipole is parallel and opposite to the direction of the dipole moment.If the observation point is far away or when the dipole is very short, then theelectric field intensity at a point on the axial line is double the electric fieldintensity at a point on the equatorial line. i.e. If l << x and l << y, then EP = 2 EQ
27. Torque on an Electric Dipole in a Uniform Electric Field:The forces of magnitude pE actopposite to each other and hence net +qforce acting on the dipole due to 2l qEexternal uniform electric field is zero.So, there is no translational motion of p qE θthe dipole. -q EHowever the forces are along differentlines of action and constitute a couple.Hence the dipole will rotate and pexperience torque. θ ETorque = Electric Force x distance t = q E (2l sin θ) t = p E sin θ Case i: If θ = 0° then t = 0. , t = pxE Case ii: If θ = 90° then t = pE ,Direction of Torque is perpendicular (maximum value).and into the plane containing p and E. Case iii: If θ = 180° then t = 0. ,SI unit of torque is newton metre (Nm).
28. Work done on an Electric Dipole in Uniform Electric Field:When an electric dipole is placed in a uniform electric field, it experiencestorque and tends to allign in such a way to attain stable equilibrium. dW = tdθ qE = p E sin θ dθ dθ + q q E 2l θ1 θ2 -q θ2 qE W = ∫ p E sin θ dθ E θ1 qE W = p E (cosθ1 - cos θ2)If Potential Energy is arbitrarily taken zero when the dipole is at 90° ,then P.E in rotating the dipole and inclining it at an angle θ isPotential Energy U = - p E cos θNote: Potential Energy can be taken zero arbitrarily at any position of thedipole. Case i: If θ = 0° then U = - pE , (Stable Equilibrium) Case ii: If θ = 90° then U = 0 , Case iii: If θ = 180° then U = pE (Unstable Equilibrium) ,
29. ELECTROSTATICS - III- Electrostatic Potential and Gauss’s Theorem1. Line Integral of Electric Field2. Electric Potential and Potential Difference3. Electric Potential due to a Single Point Charge4. Electric Potential due to a Group of Charges5. Electric Potential due to an Electric Dipole6. Equipotential Surfaces and their Properties7. Electrostatic Potential Energy8. Area Vector, Solid Angle, Electric Flux9. Gauss’s Theorem and its Proof10. Coulomb’s Law from Gauss’s Theorem11. Applications of Gauss’s Theorem: Electric Field Intensity due to Line Charge, Plane Sheet of Charge and Spherical Shell
30. Line Integral of Electric Field (Work Done by Electric Field):Negative Line Integral of Electric Field represents the work done by the electricfield on a unit positive charge in moving it from one point to another in theelectric field. B Y WAB = dW = - E . dl A FLet q0 be the test charge in place of the unit Apositive charge. rA +q0 r BThe force F = +q0E acts on the test charge rBdue to the source charge +q. +q O XIt is radially outward and tends to acceleratethe test charge. To prevent thisacceleration, equal and opposite force –q0E Zhas to be applied on the test charge.Total work done by the electric field on the test charge in moving it from A to Bin the electric field is B qq0 1 1 WAB = dW = - E . dl = 4πε0 [r B - rA ] A
31. B qq0 1 1 WAB = dW = - E . dl = 4πε0 [r B - rA ] A1. The equation shows that the work done in moving a test charge q0 from point A to another point B along any path AB in an electric field due to +q charge depends only on the positions of these points and is independent of the actual path followed between A and B.2. That is, the line integral of electric field is path independent.3. Therefore, electric field is ‘conservative field’.4. Line integral of electric field over a closed path is zero. This is another condition satisfied by conservative field. B E . dl = 0 A Note: Line integral of only static electric field is independent of the path followed. However, line integral of the field due to a moving charge is not independent of the path because the field varies with time.
32. Electric Potential:Electric potential is a physical quantity which determines the flow of chargesfrom one body to another.It is a physical quantity that determines the degree of electrification of a body.Electric Potential at a point in the electric field is defined as the work done inmoving (without any acceleration) a unit positive charge from infinity to thatpoint against the electrostatic force irrespective of the path followed. B qq0 1 1 WAB q 1 1WAB = - E . dl = 4πε0 [rB - rA ] or q0 = 4πε0 [rB - rA ] AAccording to definition, rA = ∞ and rB = r (where r is the distance from the source charge and the point of consideration) W∞B q W∞B =V V= q0 = 4πε0 r q0SI unit of electric potential is volt (V) or J C-1 or Nm C-1.Electric potential at a point is one volt if one joule of work is done in movingone coulomb charge from infinity to that point in the electric field.
33. Electric Potential Difference:Electric Potential Difference between any two points in the electric field isdefined as the work done in moving (without any acceleration) a unit positivecharge from one point to the other against the electrostatic force irrespectiveof the path followed. B qq0 1 1 WAB q 1 1WAB = - E . dl = 4πε0 [r B - rA ] or q0 = 4πε0 [rB - rA ] A WAB q 1 q 1 - = VB - VA q0 = 4πε0 rB 4πε0 rA WAB VB - VA = ∆V = q01. Electric potential and potential difference are scalar quantities.2. Electric potential at infinity is zero.3. Electric potential near an isolated positive charge (q > 0) is positive and that near an isolated negative charge (q < 0) is negative.4. cgs unit of electric potential is stat volt. 1 stat volt = 1 erg / stat coulomb
34. Electric Potential due to a Single Point Charge:Let +q0 be the test chargeplaced at P at a distance x E dx +q0 q0 Efrom the source charge +q. B ∞ +q Q PThe force F = +q0E is rradially outward and tends xto accelerate the test charge.To prevent this acceleration, equal and opposite force –q0E has to be appliedon the test charge.Work done to move q0 from P to Q through ‘dx’ against q0E isdW = F . dx = q0E . dx or dW = q0E dx cos 180° = - q 0E dx q q0 qdW = - dx E= 4πε0 x2 4πε0 x2 W∞B qTotal work done to move q0 from A to B (from ∞ to r ) is q0 = B 4πε0 r r r q q0 q q0 1 qW∞B = dW = - dx =- dx ∞ 4πε0 x2 4πε0 x2 x2 V = ∞ ∞ 4πε0 r
35. Electric Potential due to a Group of Point Charges:The net electrostatic potential at a point in the q1electric field due to a group of charges is thealgebraic sum of their individual potentials at that r1point. qn q2 +1 C r2 VP = V1 + V2 + V3 + V4 + …………+ Vn rn P r4 r3 1 n qi V= ∑ 4πε0 i=1 ri q4 q3 1 n qi ( in terms of V= ∑ position vector ) 4πε0 i=1 │ r - ri │1. Electric potential at a point due to a charge is not affected by the presence of other charges.2. Potential, V α 1 / r whereas Coulomb’s force F α 1 / r2.3. Potential is a scalar whereas Force is a vector.4. Although V is called the potential at a point, it is actually equal to the potential difference between the points r and ∞.
36. Electric Potential due to an Electric Dipole:i) At a point on the axial line: 1 qVP = q+ 4πε0 (x – l) A B +1 C 1 -qVP = -q O +q P q- 4πε0 (x + l) p l l VP = VP + VP x q+ q- q 1 1 VP = 4πε0 [ (x – l) - (x + l) ] 1 q . 2l VP = 4πε0 (x2 – l2) 1 p VP = 4πε0 (x2 – l2)
37. ii) At a point on the equatorial line: 1 q VQ = q+ 4πε0 BQ Q 1 -q VQ = q- 4πε0 AQ y A θ θ B VQ = VP + VP -q O +q q+ q- p q 1 1 l l VQ = 4πε0 [ BQ - AQ ] VQ = 0 BQ = AQ The net electrostatic potential at a point in the electric field due to an electric dipole at any point on the equatorial line is zero.
38. Equipotential Surfaces:A surface at every point of which the potential due to charge distribution isthe same is called equipotential surface.i) For a uniform electric field: E V1 V2 V3 E Plane Equipotential Surfaces + Spherical Equipotential Surfaces ii) For an isolated charge:
39. Properties of Equipotential Surfaces:1. No work is done in moving a test charge from one point to another on an equipotential surface. WAB VB - VA = ∆V = q0 If A and B are two points on the equipotential surface, then VB = VA . WAB =0 or WAB = 0 q02. The electric field is always perpendicular to the element dl of the equipotential surface. Since no work is done on equipotential surface, B WAB = - E . dl = 0 i.e. E dl cos θ = 0 A As E ≠ 0 and dl ≠ 0, cos θ = 0 or θ = 90°
40. 3. Equipotential surfaces indicate regions of strong or weak electric fields. Electric field is defined as the negative potential gradient. dV dV E=- or dr = - dr E Since dV is constant on equipotential surface, so 1 dr α E If E is strong (large), dr will be small, i.e. the separation of equipotential surfaces will be smaller (i.e. equipotential surfaces are crowded) and vice versa.4. Two equipotential surfaces can not intersect. If two equipotential surfaces intersect, then at the points of intersection, there will be two values of the electric potential which is not possible. (Refer to properties of electric lines of force) Note: Electric potential is a scalar quantity whereas potential gradient is a vector quantity. The negative sign of potential gradient shows that the rate of change of potential with distance is always against the electric field intensity.
41. Electrostatic Potential Energy:The work done in moving a charge q from infinity to a point in the fieldagainst the electric force is called electrostatic potential energy. W=qVi) Electrostatic Potential Energy Y of a Two Charges System: A (q1) 1 q1q2 r2 - r1 U = r1 4πε0 │ r2 - r1 │ B (q2) r2 or O X 1 q1q2 U= 4πε0 r12 Z
42. ii) Electrostatic Potential Energy Y C (q3) of a Three Charges System: r3 - r1 A (q1) 1 q1q2 1 q1q3 r3 - r2 U= + r1 r2 - r1 4πε0 │ r2 - r1 │ 4πε0 │ r3 - r1 │ r3 B (q2) r2 1 q2q3 + O X 4πε0 │ r3 - r2 │ Z 1 q1q2 q1q3 q2q3or U= 4πε0 [ r12 + r31 + r32 ]iii) Electrostatic Potential Energy of an n - Charges System: n qi qj U= 1 2 [ 1 4πε0 ∑ ∑ i=1 n j=1 │ rj - ri │ ] i≠j
43. Area Vector: nSmall area of a surface can be represented by a vector. dS dS = dS n dS Electric Flux: S Electric flux linked with any surface is defined as the total number of electric lines of force that normally pass through that surface. Electric flux dΦ through a small area dS dS element dS due to an electric field E at an 90° θ angle θ with dS is dS dΦ = E . dS = E dS cos θ dS Total electric flux Φ over the whole θ E surface S due to an electric field E is S Φ= E . dS = E S cos θ = E . S S Electric flux is a scalar quantity. But it is a θ property of vector field. dS SI unit of electric flux is N m2 C-1 or J m C -1.
44. Special Cases:1. For 0°< θ < 90° Φ is positive. ,2. For θ = 90° Φ is zero. ,3. For 90°< θ < 180° Φ is negative. ,Solid Angle:Solid angle is the three-dimensional equivalent of an ordinary two-dimensional plane angle.SI unit of solid angle is steradian.Solid angle subtended by area element dS at the rcentre O of a sphere of radius r is dS cos θ θ n d = dS r2 r dS cos θ = d = = 4π steradian d r2 S S
45. Gauss’s Theorem:The surface integral of the electric field intensity over any closed hypotheticalsurface (called Gaussian surface) in free space is equal to 1 / ε0 times the netcharge enclosed within the surface. 1 n ΦE = E . dS = ∑ qi ε0 i=1 SProof of Gauss’s Theorem for Spherically Symmetric Surfaces: 1 q dΦ = E . dS = r . dS n E 4πε0 r2 1 q dS dΦ = r . n r dS 4πε0 r2 O• +q r Here, r . n = 1 x 1 cos 0°= 1 1 q dS dΦ = 4πε0 r2 1 q 1 q q ΦE = dΦ = dS = 4π r2 = 4πε0 r2 4πε0 r2 ε0 S S
46. Proof of Gauss’s Theorem for a Closed Surface of any Shape: 1 q E dΦ = E . dS = r . dS n 4πε0 r2 r 1 q dS θ n dΦ = r . n dS 4πε0 r2 r Here, r . n = 1 x 1 cos θ d = cos θ +q • q dS cos θ dΦ = 4πε0 r2 q q q ΦE = dΦ = d = 4π = 4πε0 4πε0 ε0 S S
47. Deduction of Coulomb’s Law from Gauss’s Theorem:From Gauss’s law, q ΦE = E . dS = ε0 E SSince E and dS are in the same direction, r dS q O• ΦE = E dS = +q r ε0 S q or ΦE = E dS = ε0 S q q E x 4π r2 = or E= ε0 4πε0 r2If a charge q0 is placed at a point where Eis calculated, then qq0 F= which is Coulomb’s Law. 4πε0 r2
48. Applications of Gauss’s Theorem:1. Electric Field Intensity due to an Infinitely Long Straight Charged Wire: E dS C r -∞ B A +∞ dS dS Gaussian surface is aFrom Gauss’s law, E l E closed surface, q around a chargeΦE = E . dS = ε0 distribution, such S that the electric field intensity has a single E . dS = E . dS + E . dS + E . dS fixed value at every point on the surface.S A B C E . dS = E dS cos 90°+ E dS cos 90°+ E dS cos 0°= E dS = E x 2 π r lS A B C C
49. q λl = ε (where λ is the liner charge density) ε0 0 λl Ex2πrl= ε0 1 λ or E= 2 πε0 r or 1 2λ E= 4 πε0 r 1 2λ In vector form, E (r) = r 4 πε0 rThe direction of the electric field intensity is radially outward from the positiveline charge. For negative line charge, it will be radially inward.Note:The electric field intensity is independent of the size of the Gaussian surfaceconstructed. It depends only on the distance of point of consideration. i.e. theGaussian surface should contain the point of consideration.
50. 2. Electric Field Intensity due to an Infinitely Long, Thin Plane Sheet of Charge: σ dS l E E dS r C E A B dSFrom Gauss’s law, q TIP: ΦE = E . dS = ε0 The field lines remain S straight, parallel and uniformly spaced. E . dS = E . dS + E . dS + E . dS S A B C E . dS = E dS cos 0°+ E dS cos 0°+ E dS cos 90°= 2E dS = 2E x π r2 S A B C
51. q σ π r2 = (where σ is the surface charge density) ε0 ε0 σ π r2 2Exπ r2 = ε0 σ σor E= In vector form, 2 ε0 E (l) = l 2 ε0The direction of the electric field intensity is normal to the plane and awayfrom the positive charge distribution. For negative charge distribution, it willbe towards the plane.Note:The electric field intensity is independent of the size of the Gaussian surfaceconstructed. It neither depends on the distance of point of consideration northe radius of the cylindrical surface.If the plane sheet is thick, then the charge distribution will be available onboth the sides. So, the charge enclosed within the Gaussian surface will betwice as before. Therefore, the field will be twice. σ E= ε0
52. 3. Electric Field Intensity due to Two Parallel, Infinitely Long, Thin Plane Sheet of Charge:Case 1: σ1 > σ2 σ1 σ2 E1 E1 E1 Region I Region II Region III E E E ( σ1 > σ2 ) E2 E2 E2 E = E1 + E2 E = E1 - E2 E = E1 + E2 σ1 + σ2 σ1 - σ2 σ1 + σ2 E= E= E= 2 ε0 2 ε0 2 ε0
53. Case 2: + σ1 & - σ2 σ1 σ2 E1 E1 E1 Region I Region II Region III E E E ( σ1 > σ2 ) ( σ1 > σ2 ) E2 E2 E2 E = E1 - E2 E = E1 + E2 E = E1 - E2 σ1 - σ2 σ1 + σ2 σ1 - σ2 E= E= E= 2 ε0 2 ε0 2 ε0
54. Case 3: + σ & - σ σ1 σ2 E1 E1 E1 Region I Region II Region III E=0 E≠0 E=0 E2 E2 E2 E = E1 - E2 E = E1 + E2 E = E1 - E2 σ1 - σ2 σ1 + σ2 σ σ1 - σ2 E= =0 E= = E= =0 2 ε0 2 ε0 ε0 2 ε0
55. 4. Electric Field Intensity due to a Uniformed Charged This Spherical Shell: E i) At a point P outside the shell: dS rFrom Gauss’s law, •P q ΦE = E . dS = ε0 S q O• RSince E and dS are in the same direction, HOLLOW q ΦE = E dS = ε0 S q or ΦE = E dS = ……… Gaussian Surface ε0 S q q Electric field due to a uniformly E x 4π r2 = or E= charged thin spherical shell at ε0 4πε0 r2 a point outside the shell is Since q = σ x 4π R2, such as if the whole charge σ R2 were concentrated at the E= ε0 r2 centre of the shell.
56. ii) At a point A on the surface of the shell:From Gauss’s law, E q dS ΦE = E . dS = ε0 S • ASince E and dS are in the same direction, q O• R q ΦE = E dS = ε0 HOLLOW S q or ΦE = E dS = ε0 S q q E x 4π R2 = or E= ε0 4πε0 R2 Electric field due to a uniformly Since q = σ x 4π R2, σ E= charged thin spherical shell at ε0 a point on the surface of the shell is maximum.
57. iii) At a point B inside the shell:From Gauss’s law, E dS q ΦE = E . dS = B ε0 S • q O• RSince E and dS are in the same direction, r’ q ΦE = E dS = HOLLOW ε0 S q or ΦE = E dS = ε0 E S q 0 Emax E x 4π r’2 = or E= ε0 4πε0 r’2 (since q = 0 inside the Gaussian surface) E=0 O R r This property E = 0 inside a cavity is used for electrostatic shielding.
58. ELECTROSTATICS - IV - Capacitance and Van de Graaff Generator1. Behaviour of Conductors in Electrostatic Field2. Electrical Capacitance3. Principle of Capacitance4. Capacitance of a Parallel Plate Capacitor5. Series and Parallel Combination of Capacitors6. Energy Stored in a Capacitor and Energy Density7. Energy Stored in Series and Parallel Combination of Capacitors8. Loss of Energy on Sharing Charges Between Two Capacitors9. Polar and Non-polar Molecules10. Polarization of a Dielectric11. Polarizing Vector and Dielectric Strength12. Parallel Plate Capacitor with a Dielectric Slab13. Van de Graaff Generator
59. Behaviour of Conductors in the Electrostatic Field:1. Net electric field intensity in the interior of a E0 conductor is zero. When a conductor is placed in an electrostatic field, the charges (free EP electrons) drift towards the positive plate leaving the + ve core behind. At an equilibrium, the electric field due to the polarisation becomes equal to the applied Enet = 0 field. So, the net electrostatic field inside the conductor is zero.2. Electric field just outside the charged conductor is perpendicular to the surface of the conductor. E cos θ E Suppose the electric field is acting at an angle other than 90° then there will be a , θ LE component E cos θ acting along the tangent SIB at that point to the surface which will tend to POS • + q n accelerate the charge on the surface leading NOT to ‘surface current’. But there is no surface current in electrostatics. So, θ = 90°and cos 90°= 0.
60. 3. Net charge in the interior of a conductor is zero. The charges are temporarily separated. The total charge of the system is zero. q ΦE = E . dS = ε 0 S Since E = 0 in the interior of the conductor, therefore q = 0.4. Charge always resides on the surface of a conductor. Suppose a conductor is given some excess charge q. Construct a Gaussian surface just q q inside the conductor. Since E = 0 in the interior of the conductor, therefore q = 0 inside the conductor. q=05. Electric potential is constant for the entire conductor. dV = - E . dr Since E = 0 in the interior of the conductor, therefore dV = 0. i.e. V = constant
61. 6. Surface charge distribution may be different at different points. q σ= S σmin σmax Every conductor is an equipotential volume (three- dimensional) rather than just an equipotential surface (two- dimensional).Electrical Capacitance: The measure of the ability of a conductor to store charges is known as capacitance or capacity (old name). q q α V or q = C V or C= V If V = 1 volt, then C = q Capacitance of a conductor is defined as the charge required to raise its potential through one unit. SI Unit of capacitance is ‘farad’ (F). Symbol of capacitance: Capacitance is said to be 1 farad when 1 coulomb of charge raises the potential of conductor by 1 volt. Since 1 coulomb is the big amount of charge, the capacitance will be usually in the range of milli farad, micro farad, nano farad or pico farad.
62. Capacitance of an Isolated Spherical Conductor:Let a charge q be given to the sphere whichis assumed to be concentrated at the centre.Potential at any point on the surface is r O• q +q V = 4πε0 r q C= V C = 4πε0 r1. Capacitance of a spherical conductor is directly proportional to its radius.2. The above equation is true for conducting spheres, hollow or solid.3. IF the sphere is in a medium, then C = 4πε0εr r.4. Capacitance of the earth is 711 µF.
63. Principle of Capacitance: A BStep 1: Plate A is positively charged and B is neutral.Step 2: When a neutral plate B is brought near A,charges are induced on B such that the side near A isnegative and the other side is positive.The potential of the system of A and B in step 1 and 2remains the same because the potential due to positiveand negative charges on B cancel out. Potential = VStep 3: When the farther side of B is earthed thepositive charges on B get neutralised and B is left onlywith negative charges. A BNow, the net potential of the system decreases due tothe sum of positive potential on A and negativepotential on B.To increase the potential to the same value as was instep 2, an additional amount of charges can be given toplate A. Potential = VThis means, the capacity of storing charges on A Potential Eincreases. decreases to vThe system so formed is called a ‘capacitor’.
64. Capacitance of Parallel Plate Capacitor:Parallel plate capacitor is an arrangement of twoparallel conducting plates of equal area σ E σseparated by air medium or any other insulatingmedium such as paper, mica, glass, wood, A Aceramic, etc. σ V=Ed= d ε0 qd or V= A ε0 q A ε0 But C= C= d V dIf the space between the plates is filled with dielectric medium of relativepermittivity εr, then A ε0 εr C= dCapacitance of a parallel plate capacitor is(i) directly proportional to the area of the plates and(ii) inversely proportional to the distance of separation between them.
65. Series Combination of Capacitors: C1 C2 C3In series combination, q q qi) Charge is same in each capacitorii) Potential is distributed in inverse V1 V2 V3 proportion to capacitances i.e. V = V1 + V2 + V3 V q q q q But V = , V1 = , V2 = and V3 = C C1 C2 C3 q q q q (where C is the equivalent capacitance or = + + effective capacitance or net capacitance or C C1 C2 C3 total capacitance) 1 1 1 1 n 1 1or = + + = ∑ C C1 C2 C3 C i=1 Ci The reciprocal of the effective capacitance is the sum of the reciprocals of the individual capacitances. Note: The effective capacitance in series combination is less than the least of all the individual capacitances.
66. Parallel Combination of Capacitors: C1In parallel combination, V q1i) Potential is same across each capacitorii) Charge is distributed in direct proportion to C2 capacitances V q2 i.e. q = q1 + q2 + q3 But q1 = C1 V , q2 = C2 V , q3 = C3 V and q = C V C3 V q3 C V = C1V + C2 V + C3 V (where C is the equivalent capacitance) n or C = C1 + C2 + C3 V C = ∑ Ci i=1The effective capacitance is the sum of the individual capacitances.Note: The effective capacitance in parallel combination is larger than thelargest of all the individual capacitances.
67. Energy Stored in a Capacitor:The process of charging a capacitor isequivalent to transferring charges from oneplate to the other of the capacitor.The moment charging starts, there is a potentialdifference between the plates. Therefore, totransfer charges against the potential differencesome work is to be done. This work is stored aselectrostatic potential energy in the capacitor.If dq be the charge transferred against thepotential difference V, then work done is dU = dW = V dq V q = dq CThe total work done ( energy) to transfer charge q is q q 1 q2 1 1 U= dq or U= or U= C V2 or U = qV C 2 C 2 2 0
68. Energy Density: 1 A ε0 U= C V2 But C= and V=Ed 2 d 1 U 1 1 U= ε0 Ad E2 or = ε0 E2 or U = ε0 E2 2 Ad 2 2 SI unit of energy density is J m-3. Energy density is generalised as energy per unit volume of the field.Energy Stored in a Series Combination of Capacitors: 1 1 1 1 1 = + + + ………. + C C1 C2 C3 Cn 1 q2 1 1 1 1 1 U= U= 2 q2 [ C1 + C2 + C3 + ………. + Cn ] 2 C U = U1 + U2 + U3 + ………. + Un The total energy stored in the system is the sum of energy stored in the individual capacitors.
69. Energy Stored in a Parallel Combination of Capacitors: C = C1 + C2 + C3 + ……….. + Cn 1 1 U= C V2 U= V2 ( C1 + C2 + C3 + ……….. + Cn ) 2 2 U = U1 + U2 + U3 + ………. + UnThe total energy stored in the system is the sum of energy stored in theindividual capacitors.Loss of Energy on Sharing of Charges between the Capacitors inParallel:Consider two capacitors of capacitances C1, C2, charges q1, q2 andpotentials V1,V2.Total charge after sharing = Total charge before sharing (C1 + C2) V = C1 V1 + C2 V2 C1 V1 + C2 V2 V= C1 + C2
70. The total energy before sharing is 1 1 C1 V1 2 C2 V22 Ui = + 2 2 The total energy after sharing is 1 Uf = (C1 + C2) V2 2 C1 C2 (V1 – V2)2 Ui– Uf = 2 (C1 + C2) Ui – Uf > 0 or Ui > UfTherefore, there is some loss of energy when two charged capacitors areconnected together.The loss of energy appears as heat and the wire connecting the two capacitorsmay become hot.
71. Polar Molecules:A molecule in which the centre of positive charges does Onot coincide with the centre of negative charges is calleda polar molecule. 105°Polar molecule does not have symmetrical shape. p H HEg. H Cl, H2 O, N H3, C O2, alcohol, etc.Effect of Electric Field on Polar Molecules: E=0 E p=0 p In the absence of external electric When electric field is applied, the field, the permanent dipoles of the dipoles orient themselves in a molecules orient in random regular fashion and hence dipole directions and hence the net dipole moment is induced. Complete moment is zero. allignment is not possible due to thermal agitation.
72. Non - polar Molecules:A molecule in which the centre of positive charges coincides with the centre ofnegative charges is called a non-polar molecule.Non-polar molecule has symmetrical shape.Eg. N2 , C H4, O2, C6 H6, etc.Effect of Electric Field on Non-polar Molecules: E=0 E p=0 pIn the absence of external When electric field is applied, the positiveelectric field, the effective charges are pushed in the direction of electricpositive and negative centres field and the electrons are pulled in thecoincide and hence dipole is direction opposite to the electric field. Due tonot formed. separation of effective centres of positive and negative charges, dipole is formed.
73. Dielectrics:Generally, a non-conducting medium or insulator is called a ‘dielectric’.Precisely, the non-conducting materials in which induced charges are producedon their faces on the application of electric fields are called dielectrics.Eg. Air, H2, glass, mica, paraffin wax, transformer oil, etc.Polarization of Dielectrics:When a non-polar dielectric slab issubjected to an electric field, dipolesare induced due to separation ofeffective positive and negative centres.E0 is the applied field and Ep is theinduced field in the dielectric.The net field is EN = E0 – Ep E=0 E0 Epi.e. the field is reduced when adielectric slab is introduced.The dielectric constant is given by E0 K= E0 - Ep
74. Polarization Vector:The polarization vector measures the degree of polarization of the dielectric. Itis defined as the dipole moment of the unit volume of the polarized dielectric.If n is the number of atoms or molecules per unit volume of the dielectric, thenpolarization vector is P=npSI unit of polarization vector is C m-2. Dielectric Dielectric strength (kV /Dielectric Strength: mm)Dielectric strength is the maximum Vacuum ∞value of the electric field intensity Air 0.8 – 1that can be applied to the dielectricwithout its electric break down. Porcelain 4–8Its SI unit is V m-1. Pyrex 14 Paper 14 – 16Its practical unit is kV mm-1. Rubber 21 Mica 160 – 200
75. Capacitance of Parallel Plate Capacitor with Dielectric Slab: V = E0 (d – t) + EN t E0 E0 K= or EN = EN K E0 Ep t d EN = E0 - Ep E0 V = E0 (d – t) + t K t V = E0 [ (d – t) + K ] A ε0 σ qA or C= But E0 = = t ε0 ε0 d 1– [ d (1 - t )] q K and C= V C0 or C= A ε0 t C= [1 – d (1 - t )] t K [ (d – t) + K ] C > C0. i.e. Capacitance increases with introduction of dielectric slab.
76. If the dielectric slab occupies the whole space between the plates, i.e. t = d,then C = K C0 CDielectric Constant K= C0 WITH DIELECTRIC SLAB Physcial Quantity With Battery With Battery disconnected connected Charge Remains the same Increases (K C0 V0) Capacitance Increases (K C0) Increases (K C0) Electric Field Decreases Remains the same EN = E0 – Ep Potential Difference Decreases Remains the same Energy stored Remains the same Increases (K U0)
77. Van de Graaff Generator: S P2 C2 S – Large Copper sphere D C1, C2 – Combs with sharp points P1, P2 – Pulleys to run belt HVR – High Voltage Rectifier M – Motor T IS – Insulating Stand C1 I S D – Gas Discharge Tube T - Target HVR P1 M
78. Principle:Consider two charged conducting spherical shells such that one issmaller and the other is larger. When the smaller one is kept inside thelarger one and connected together, charge from the smaller one istransferred to larger shell irrespective of the higher potential of the largershell. i.e. The charge resides on the outer surface of the outer shell andthe potential of the outer shell increases considerably.Sharp pointed surfaces of a conductor have large surface chargedensities and hence the electric field created by them is very highcompared to the dielectric strength of the dielectric (air).Therefore air surrounding these conductors get ionized and the likecharges are repelled by the charged pointed conductors causingdischarging action known as Corona Discharge or Action of Points. Thesprayed charges moving with high speed cause electric wind.Opposite charges are induced on the teeth of collecting comb (conductor)and again opposite charges are induced on the outer surface of thecollecting sphere (Dome).
79. Construction: Van de Graaff Generator consists of a large (about a few metres inradius) copper spherical shell (S) supported on an insulating stand (IS)which is of several metres high above the ground. A belt made of insulating fabric (silk, rubber, etc.) is made to run overthe pulleys (P1, P2 ) operated by an electric motor (M) such that it ascendson the side of the combs. Comb (C1) near the lower pulley is connected to High Voltage Rectifier(HVR) whose other end is earthed. Comb (C2) near the upper pulley isconnected to the sphere S through a conducting rod. A tube (T) with the charged particles to be accelerated at its top andthe target at the bottom is placed as shown in the figure. The bottom endof the tube is earthed for maintaining lower potential. To avoid the leakage of charges from the sphere, the generator isenclosed in the steel tank filled with air or nitrogen at very high pressure(15 atmospheres).
80. Working:Let the positive terminal of the High Voltage Rectifier (HVR) isconnected to the comb (C1). Due to action of points, electric wind iscaused and the positive charges are sprayed on to the belt (silk orrubber). The belt made ascending by electric motor (EM) and pulley(P1) carries these charges in the upward direction.The comb (C2) is induced with the negative charges which arecarried by conduction to inner surface of the collecting sphere(dome) S through a metallic wire which in turn induces positivecharges on the outer surface of the dome.The comb (C2) being negatively charged causes electric wind byspraying negative charges due to action of points which neutralizethe positive charges on the belt. Therefore the belt does not carryany charge back while descending. (Thus the principle ofconservation of charge is obeyed.) Contd..
81. The process continues for a longer time to store more and morecharges on the sphere and the potential of the sphere increasesconsiderably. When the charge on the sphere is very high, theleakage of charges due to ionization of surrounding air alsoincreases.Maximum potential occurs when the rate of charge carried in bythe belt is equal to the rate at which charge leaks from the shelldue to ionization of air.Now, if the positively charged particles which are to beaccelerated are kept at the top of the tube T, they get accelerateddue to difference in potential (the lower end of the tube isconnected to the earth and hence at the lower potential) and aremade to hit the target for causing nuclear reactions, etc.
82. Uses:Van de Graaff Generator is used to produce very highpotential difference (of the order of several million volts) foraccelerating charged particles. The beam of accelerated charged particles are used to trigger nuclear reactions. The beam is used to break atoms for various experiments in Physics. In medicine, such beams are used to treat cancer.It is used for research purposes.