Electrostatics PDF

8. Electrostatics charge configurations with the help of some Can you recall? examples. 1. What are conservative forces? 8.2.1 Electric Field Inte...

8. Electrostatics charge configurations with the help of some Can you recall? examples. 1. What are conservative forces? 8.2.1 Electric Field Intensity due to Uniformly 2. What is potential energy ? Charged Spherical Shell or Hollow Sphere: 3. What is Gauss’ law and what is a Consider a sphere of radius R with its Gaussian surface? centre at O, charged to a uniform surface charge density σ (C/m2 ) placed in a dielectric 8.1 Introduction: medium of permittivity ε (ε = ε 0 k ). The total In XIth Std we have studied the Gauss’ charge on the sphere, q = σ × 4πR2 Law which gives the relationship between By Gauss’ theorem, the net flux through a the electric charge and its electric field. It closed Gaussian surface also provides equivalent methods for finding φ = q/ε0 (for air/vacuum k=1) electric field intensity by relating values of where q is the total charge inside the closed the field at a closed Gaussian surface and the surface. total charge enclosed by it. It is a powerful tool which can be applied for the calculation of the electric field when it originates from charge distribution of sufficient symmetry. The Gauss' law is written as   q    E  ds  --- (8.1) 0 where φ is the total flux coming out of a closed surface and q is the total charge inside Fig. 8.1: Uniformly charged spherical shell or the closed surface. hollow sphere. To find the electric field intensity at Common steps involved in calculating electric field intensity by using Gauss’ law: a point P, at a distance r from the centre of 1. Identify the charge distribution as the charged sphere, imagine a concentric linear/cylindrical/spherical charge density. Gaussian sphere of radius r passing through 2. Visualize a Gaussian surface justifying its P. Let ds be a small area around the point P symmetry for the given charge distribution. on the Gaussian surface. Due to symmetry and 3. Obtain the flux by Gauss’ law (Let this be spheres being concentric, the electric field at Eq. (A)) each point on the Gaussian surface has the 4. With the electric field intensity E same magnitude E and it is directed radially as unknown, obtain electric flux by outward. Also, the angle between the direction calculation, using geometry of the structure of E and the normal to the surface of the sphere and symmetry of the Gaussian surface (Let (ds) is zero i.e., cos θ = 1 this be Eq. (B))   ∴ E. ds = E ds cos θ = E ds 5. Equate RHS of Eq. (A) and Eq. (B) and ∴ flux d φ through the area ds = E ds calculate E. Total electric flux  through the Gaussian  8.2 Application of Gauss' Law: surface    E ds   Eds  E  ds In this section we shall see how to obtain ∴ φ = E 4π r 2 --- (8.2) the electric field intensity for some symmetric From equations (8.1) and (8.2), 186  q/ε0 = E 4π r2 ∴ E = q/ 4 0 r 2 --- (8.3) Since q = σ × 4πR 2 We have E = σ × 4πR2 / 4 0 r 2 ∴E = σR2 / ε 0 r 2 --- (8.4) From Eqn. (8.3) it can be seen that, the electric field at a point outside the shell is the same as that due to a point charge. Thus it can be concluded that a uniformly charged sphere is equivalent to a point charge at its center. Fig. 8.2: Infinitely long straight charged wire Case (i) If point P lies on the surface of the (cylinder). charged sphere: r = R To find the electric field intensity at P ,at ∴ E= q/ 4 0 R 2 =σ/ ε 0 a distance r from the axis of the charged wire, Case (ii) If point P lies inside the sphere: Since imagine a coaxial Gaussian cylinder of length l and radius r (closed at each end by plane caps there are no charges inside σ = 0, normal to the axis) passing through the point P. ∴ E = 0. Consider a very small area ds at the point P on Example : 8.1 the Gaussian surface. A sphere of radius 10 cm carries a charge of By symmetry, the magnitude of the electric 1µC. Calculate the electric field field will be the same at all the points on the (i) at a distance of 30 cm from the center curved surface of the cylinder and will be of the sphere directed radially outward. The angle between (ii) at the surface of the sphere and the direction of E and the normal to the curved or (iii)  at a distance of 5 cm from the center of flat surface of the cylinder (ds) is zero or (π/2) i.e., the sphere. cos θ =1 or cos (π/2) = 0. ∴ E.ds = Eds cos θ = Eds Solution: Given: q = 1µC = 1 × 10-6 C Flux d f through the area ds = E ds. (i) Electric intensity at a distance r is Total electric flux through the Gaussian E = q/ 4 0 r 2 cylindrical  surface For r = 30 cm = 0.3 m    E  ds   Eds  E  ds  9  109  1 10 6 ∴ f = E. 2πrl --- (8.5) E =  = 105 N/C  0.3 2 From equations (8.1) and (8.5) (ii) E on the surface of the sphere, R =10 q/ ε 0 = E 2π rl cm = 0.10m Since λ = q/l , q = λ l E = q/ 4 0 R 2 ∴ λ l / ε 0 = E 2π rl E = λ / 2 0 r --- (8.6) 9  109  1 10 6 The direction of the electric field E is = = 9 × 105 N/C  0.10  2 directed outward if λ is positive and inward if (iii) E at a point 5 cm away from the centre is λ negative (Fig 8.3). i.e. inside the sphere E = 0. 8.2.2 Electric Field Intensity due to an Infinitely Long Straight Charged Wire: Consider a uniformly charged wire of infinite length having a constant linear charge density λ (charge per unit length), kept in a medium of permittivity ε (ε = ε 0 k ). Fig. 8.3: Direction of the field for two types of charges. 187  Example 8.2: The length of a straight thin By symmetry the electric field is at right wire is 2 m. It is uniformly charged with a angles to the end caps and away from the positive charge of 3µC. Calculate plane. Its magnitude is the same at P and P'. (i) the charge density of the wire The flux passing through the curved surface is (ii) the electric intensity due to the wire at zero as the electric field is tangential to this a point 1.5 m away from the center of the surface. wire ∴ the total flux through the closed Gaussian Solution: Given surface is given by     Eds     Eds     Eds  charge q = 3 µC = 3 × 10-6 C P P' curved surface Length l = 2 m, r = 1.5 m (since θ = 0, cos θ =1) (i) Charge Density λ = Charge/ length = EA + EA 3 10 6 φ ∴ = 2EA --- (8.7) = = 1.5 × 10-6 C m-1 2 If σ is the surface charge density then (ii) Electric Intensity E = λ / 2 0 r σ = q/A, q = σA 1.5  10 6 ∴ Eq. (8.1) can be written as = 2  3.142  8.85  10 12  1.5 φ = σA/ ε 0  --- (8.8)  = 1.798 × 104 N C-1 From Eq. (8.7) and Eq. (8.8) 8.2.3 Electric Field due to a Charged Infinite 2EA = σA/ ε 0 ∴ E = σ/2 ε 0 Plane Sheet: Example: 8.3 The charge per unit area Consider a uniformly charged infinite of a large flat sheet of charge is 3µC/m2. plane sheet with surface charge density σ. By symmetry electric field is perpendicular Calculate the electric field intensity at to plane sheet and directed outwards ,having a point just near the surface of the sheet, same magnitude at a given distance on either measured from its midpoint. sides of the sheet. Let P be a point at a distance Solution: Given r from the sheet and E be the electric field at P. Surface Charge Density = σ = 3× 10-6 Cm-2 Electric Intensity E = σ/2 ε 0 3 10 6 = = 1.695 × 105 N C-1 2  8.85  10 12 Can you recall? What is gravitational Potential ? 8.3 Electric Potential and Potential Energy: Fig. 8.4: Charged infinite plane sheet. We have studied earlier that the potential To find the electric field due to a charged energy of a system is the stored energy that infinite plane sheet at P, we consider a depends upon the relative positions of its Gaussian surface around P in the form of a constituents. Electrostatic potential energy is cylinder having cross sectional area A and length 2r with its axis perpendicular to the the work done against the electrostatic forces plane sheet. The plane sheet passes through to achieve a certain configuration of charges the middle of the length of the cylinder such in a given system. Since every system tries that the ends of the cylinder (called end caps to attain the lowest potential energy, work P and P') are equidistant (at a distance r) from is always required to be done to change the the plane sheet. configuration. 188  We know that like charges repel and   1  Qq0 unlike charges attract each other. A charge FE =    2 r̂ exerts a force on any other charge in its  4  0  r vicinity. Some work is always done to move where r is the unit vector in the direction of   a charge in the presence of another charge. r. Negative sign shows r and FE are Thus, any collection of charges possesses oppositely directed. potential energy. Consider a positive charge ∴For a system of two point charges, Q fixed at some point in space. For bringing B r2  1  Qq0  any other positive charge close to it, work is ΔU = ∫dU =     2 rˆ.dr  necessary. This work is equal to the change A r1  4  0  r r in the potential energy of the system of two  1   1  2 charges.  U     Qq0    4  0   r r1 Thus, work done against a electrostatic force = Increase in the potential  1   1 1   energy of the system.   Qq0    ∴ F. dr = dU,  4  0   r2 r1  where dU is the increase in potential energy  The change in the potential energy depends when  the charge is displaced through dr and only upon the end points and is independent of F is the force exerted on the charge. Expression for potential energy: the path taken by the charge. The change in Let us consider the electrostatic field due potential energy is equal to the work done WAB to a source charge +Q placed at the origin O. against the electrostatic force. Let a small charge + q0  be brought from point  1   1 1 WAB  U    Qq0    A to point B at respective distances r1 and r2  4  0   r2 r1  from O, against the repulsive forces on it. So far we have defined/calculated the change in the potential energy for system of charges. It is convenient to choose infinity to be the point of zero potential energy as the Fig. 8.5: Change +q0 displaced by dr towards electrostatic force is zero at r  . charge +Q. Thus, the potential energy U of the system  Work done against the electrostatic force of two point charges q1 and q2 separated by r FE , in displacing the charge q0 through a can be obtained from the above equation by  small displacement dr appears as an increase using r1   andr2  r. It is then given by in the potential  energy of the system.  1   q1q2 .  U r    dU = FE dr = - FE.dr   --- (8.9) Negative sign appears because the  4  0   r   Units of potential energy : displacement  d r is against the electrostatic force FE. SI unit= joule (J) For the displacement of the charge from “One joule is the energy stored in moving the initial position A to the final position B, a charge of 1C through a potential difference the change in potential energy ΔU, can be of 1 volt. Another convenient unit of energy is obtained by integrating dU electron volt (eV), which is the change in the B B   kinetic energy of an electron while crossing ∴ΔU = ∫ dU =   FE.dr A A two points maintained at a potential difference The electrostatic force (Coulomb force) of 1 volt.” between the two charges separated by distance 1 eV = 1.6 × 10-19 joule r is Other related units are: 189  1 meV = 1.6 × 10-22 J the work done on a unit positive charge, dw = 1 kev = 1.6 × 10-16 J dV = difference in potential between M and N. Concept of Potential:  dV   Edx Equation (8.9) gives the potential energy dV E of a two particles system at a distance r from dx each other. Thus the electric field at a point is the  1   q1q2  U r      negative gradient of the potential at that point.  4  0 r   r  Zero potential:  q   q  The nature of potential is such that its   1  q2   2  q1  4  0 r   4  0 r  zero point is arbitrary. This does not mean  q  that the choice of zero point is insignificant. The quantity V  r     depends upon Once the zero point of the potential is set,  4  0 r  then every potential is measured with respect the charge q and location of a point at a distance to that reference. The zero potential is set r from it. This is defined as the electrostatic conveniently. potential of the charge q at a distance r from it. In case of a point charge or localised In terms of potential, we can write the collection of charges, the zero point is set at potential energy of the system of two charges infinity. For electrical circuits the earth is as U  r   V1  r  q2  V2  r  q1 , usually taken to be at zero potential. where V1(r) and V2(r) are the respective Thus the potential at a point A in an potentials of charges q1 and q2 at distance r electric field is the amount of work done to from either. bring a unit positive charge from infinity to ∴ Electrostatic potential energy (U ) = electric point A. potential V × charge q Or, Electrostatic Potential (V ) = Electrostatic Example 8.4: Potential at a point A in Potential Energy per unit charge. space is given as 4 × 105 V. i.e., V = U /q (i) Find the work done in bringing a charge Electrostatic potential difference between any of 3 µC from infinity to the point A. two points in an electric field can be written as (ii) Does the answer depend on the path U –U V2 − V1 = 2  1 = dW = work done dW along which the charge is brought ? q q Solution : Given (or change in PE) per unit charge to move the Potential (V ) at the point A = 4 × 105 V charge from point 2 to point 1. Charge q0 = 3 µC =3× 10 –6 C Relation between electric field and electric (i) Work done in bringing the charge from potential: infinity to the point A is Consider the electric field produced by a W∞ = q0 V charge +q kept at point O (see Fig. 8.6). Let us = 3 ×10-6 × 4 × 105 calculate the work done to move a unit positive = 12 × 10-1 charge from point M to point N which is at a W∞ = 1.2 J small distance dx from M. The direction of  (ii) No, the work done is independent of the electric field at M is along OM. Thus the the path. force acting on the unit positive charge is along  Example 8.5 If 120 µ J of work is done in OM. The work done = dW = - Fdx = -Edx. The carrying a charge of 6 µ C from a place negative sign indicates that we are moving the where the potential is 10 volt to another charge against the force acting on it. As it is 190  r place where the potential is V, find V -q x 2 = dx Solution: Given : WAB = 120 µJ , 4 0  q0 = 6 µC, VA = 10 V, VB = V -q  1  r  1     x 2 dx   As VB - VA = AB W 4 0  x   x  q0 q 1 1   1  120  10 6 J       0  V - (10) = 4 0  r      6  10 6 C q V - (10) = 20 W = --- (8.12) ∴ V = 30 volt 4 0 r By definition this is the electrostatic 8.4 Electric Potential due to a Point Charge, potential at A due to charge q. a Dipole and a System of Charges: q ∴V = W =  --- (8.13) a) Electric potential due to a point charge: 4 0 r Here, we shall derive an expression for the A positively charged particle produces electrostatic potential due to a point charge. a positive electric potential and a negatively Figure 8.6 shows a point charge +q, charged particle produces a negative electric located at point O. We need to determine its potential q potential at a point A, at a distance r from it. At r = ∞, V = 0  This shows that the electrostatics potential is zero at infinity. Equation (8.13) shows that for any point Fig. 8.6: Electric potential due to a point charge. at a distance r from the point charge q, the As seen above the electric potential at a value of V is the same and is independent of point A is the amount of work done per unit the direction of r. Hence electrostatic potential positive charge, which is displaced from ∞ to due to a single charge is spherically symmetric. point A. As the work done is independent of Figure 8.7 shows how electric potential the path, we choose a convenient path along 1 1 ( V α ) and electric field (E α 2 ) vary with the line extending OA to ∞. r r r, the distance from the charge. Let M be an intermediate point on this path where OM = x. The electrostatic force on a unit positive charge at M is of magnitude 1 q F   2 --- (8.10) 4  0 x Fig. 8.7: Variation It is directed away from O, along OM. For of electric field infinitesimal displacement dx from M to N, the and potential with distance amount of work done is given by ∴dW = - Fdx --- (8.11) The negative sign appears as the displacement is directed opposite to that of the force. Remember this ∴ Total work done in displacing the unit positive charge from ∞ to point A is given by Due to a single charge at a distance r, r r 1 q Force (F) α 1 / r 2 , Electric field (E) α 1/ r 2 W    Fdx    dx   4 0 x 2 but Potential (V) α 1/ r. 191  Example 8.6: A wire is bent in a circle of –q V2 = radius 10 cm. It is given a charge of 250µC 4 0 r2 which spreads on it uniformly. What is the The electrostatic potential is the work electric potential at the centre ? done by the electric field per unit charge,  W Solution : Given : V  .  Q q = 250 µC = 250 × 10-6 C The potential at C due to the dipole is, R = 10 cm = 10-1 m q 1 1 V = ? VC  V1  V2   -  1 q 9 109  250 10 6  4 0  r1 r2  As V = = 4π  0 r 10 1 By geometry, 2 r 12  r    2 r  cos  2 2 = 2.25 × 107 volt r 2  r    2 r  cos  2 2 b) Electric potential due to an electric dipole: We have studied electric and magnetic r 2  r 2  1    2  cos   2 dipoles in XIth Std. Figure 8.8 shows an 1    r2 r  electric dipole AB consisting of two charges r  r 2  1    2  cos   2 2 +q and -q separated by a finite distance 2ℓ. 2   r2 r  Its dipole moment is p, of magnitude p = q × 2l, directed from -q to +q. The line joining the For a short dipole, 2  >  r is small ∴ 2 can be neglected A straight line drawn perpendicular to the axis r and passing through centre O of the electric  r 1  r 2  1  2  cos   2 dipole is called equator of dipole.  r  In order to determine the electric potential 2  2  due to a dipole, let the origin be at the centre r 2  r 2 1  cos    r  (O) of the dipole. 1  r  r  1  2 cos   2 1  r    1 r2  r  1  2 cos   2  r  1 1 1  2  2   1  cos   and r1 r  r  1 Fig. 8.8: Electric potential due to an electric 1 1  2  2  1  cos   dipole. r2 r  r  Let C be any point near the electric dipole  1 q 1 2  cos   2 VC  V1  V2   1  4 0  r  at a distance r from the centre O inclined at an r  angle θ with axis of the dipole. r1 and r2 are  the distances of point C from charges +q and 1 2  cos   1 2  -q, respectively.  1   r r  Potential at C due to charge +q at A is,  V = + q Using binomial expansion, ( 1 + x)n = 1 + 4 0 r1 1 nx, x C1 ∴V = + + V - V1 C1 C2 C3 Thus capacitance of metal plate P1, is increased by placing an identical earth connected metal plate P2 near it. Such an arrangement is called capacitor. It is symbolically shown as. ⊥ ⊥ If the conductors are plane sheets then it is called parallel plate capacitor. We also have Fig. 8.26: Effective capacitance of three spherical capacitor, cylindrical capacitor etc. capacitors in series. 204  Let Cs represent the equivalent capacitance In this combination all the capacitors shown in Fig. 8.26, then V = Q have the same potential difference but the CS plate charges (± Q1) on capacitor1, (± Q2) Q Q Q Q on the capacitor 2 and (± Q3) on capacitor 3 ∴ = + + CS C1 C 2 C3 are not necessarily the same. If charge Q is 1 1 1 1 applied at point A then it will be distributed to ∴ = + + Cs C1 C2 C3 the capacitors depending on the capacitances. ( for 3 capacitors in series) ∴Total charge Q can be written as Q = Q1 + This argument can be extended to yield Q2 + Q3 = C1 V + C2 V + C3V an equivalent capacitance for n capacitors Let Cp be the equivalent capacitance of connected in series. The reciprocal of the combination then Q = CpV equivalent capacitance is equal to the sum of ∴C pV = C1V + C2 V + C3 V the reciprocals of individual capacitances of ∴ Cp = C1 + C2 + C3 the capacitors. The general formula for effective 1 1 1 1    .................  capacitance Cp for parallel combination of n Ceq C1 C2 Cn capacitors follows similarly If all capacitors are equal then Cp = C1 + C2+.............. + Cn 1 n C If all capacitors are equal then Ceq = nC = = or Ceq Ceq C n Remember this Remember this Capacitors are combined in parallel when Series combination is used when a we require a large capacitance at small potentials. high voltage is to be divided on several capacitors. Capacitor with minimum Example 8.15 When 108 electrons are capacitance has the maximum potential transferred from one conductor to another, a difference between the plates. potential difference of 10 V appears between b) Capacitors in Parallel: the conductors. Find the capacitance of the The parallel arrangement of capacitors two conductors. is as shown in Fig. 8.27 below, where the Solution : Given : insulated plates are connected to a common Number of electrons n = 108 terminal A which is joined to the source of V = 10 volt potential, while the other plates are connected ∴charge transferred to another common terminal B which is Q = ne = 108 × 1.6 × 10-19 earthed. (∵ e = 1.6 × 10-19 C) = 1.6 × 10-11 C ∴ Capacitance between two conductors Q 1.6  10 11 C= = = 1.6 × 10-12 F V 10 Example 8.16: From the figure given below find the value of the capacitance C if the equivalent capacitance between A and B is to be 1 µF. All other capacitors are in micro Fig. 8.27: Parallel combination of capacitors. farad. 205  A parallel plate capacitor consists of two thin conducting plates each of area A, held parallel to each other, at a suitable distance d apart. One of the charged plates is isolated and charged and the other is earthed as shown in Fig. 8.28. Solution : Given : C1 = 8 µF , C2 = 4 µF , C3 = 1µF , C4 = 4 µF , C5 = 4 µF The effective capacitance of C4 and C5 in parallel = C4 + C5 = 4 + 4 = 8 µF The effective capacitance of C3 and 8 µF in series 18 8 = = µF Fig. 8.28: Capacitor with dielectric. 1 8 9 When a charge +Q is given to the isolated The capacitance 8 µF is in parallel with plate, then a charge -Q is induced on the inner the series combination of C1 and C2. Their face of earthed plate and +Q is induced on effective combination is its outer face. But as this face is earthed the C1C2 8 8 ×4 8 32 + ⇒ + ⇒ µF charge +Q being free, flows to earth. C1 +C2 9 12 9 9 32 In the outer regions the electric fields due This capacitance of µF is in series with to the two charged plates cancel out. The net 9 C and their effective capacitance is given to field is zero. be 1µF   E= - =0 32 2 0 2 0 C 9 1 In the inner regions between the two 32 capacitor plates the electric fields due to the  C 9 two charged plates add up. The net field is thus 32 32  C   C    Q 9 9 E= + = = --- (8.20) = 1.392 µF 2 0 2 0 0 A 0 The direction of E is from positive to 8.10 Capacitance of a Parallel Plate negative plate. Capacitor Without and With Dielectric Let V be the potential difference between Medium Between the Plates: the 2 plates. Then electric field between the In section 8.8 we have studied the plates is given by behaviour of dielectrics in an external field. Let V E = or V = Ed --- (8.21) us now see how the capacitance of a parallel d plate capacitor is modified when a dielectric is Substituting Eq. (8.20) in Eq. (8.21) we Q introduced between its plates. get V = d Aε 0 a) Capacitance of a parallel plate capacitor Capacitance of the parallel plate capacitor without a dielectric: is given by 206  Let E0 be the electric field intensity Remember this between the plates before the introduction of the dielectric slab. Then the potential difference (1) If there are n parallel plates then there between the plates is given by V0 = E0d, will be (n-1) capacitors, hence  Q Aε 0 where Eo   , and C = (n - 1)  o A o d (2) For a spherical capacitor, consisting σ is the surface charge density on the plates. of two concentric spherical conducting Let a dielectric slab of thickness t (t < d) be shells with inner and outer radii as a and b introduced between the plates of the capacitor. respectively, the capacitance C is given by The field E0 polarizes the dielectric, inducing  ab  charge - Qp on the left side and +Qp on the right C = 4 0   side of the dielectric as shown in Fig. 8.29. b-a  These induced charges set up a field Ep (3) For a cylindrical capacitor, consisting inside the dielectric in the opposite direction of of two coaxial cylindrical shells with radii E0. The induced field is given by of the inner and outer cylinders as a and b, and length ℓ, the capacitance C is given by  Q  Qp  Ep  p  p  p   2 0   o A o  A  C b loge The net field (E) inside the dielectric a reduces to E0- Ep. Q Q A Hence, C= = = 0 E  Eo  V  Qd  d E = Eo - Ep = o  =k , ---   (8.22) k  Eo - Ep   A 0  where k is a constant called the dielectric b) Capacitance of a parallel plate capacitor constant. Q with a dielectric slab between the plates: E  or Q  Ak  0 E --- (8.23) Let us now see how Eq. (8.22) gets A 0 k modified with a dielectric slab in between the Remember this plates of the capacitor. Consider a parallel plate capacitor with the two plates each of area The dielectric constant of a conductor is A separated by a distance d. The capacitance infinite. of the capacitor is given by The field Ep exists over a distance t and E0 Aε 0 over the remaining distance (d - t) between the C0 = d capacitor plates. Hence the potential difference between the capacitor plates is V = Eo  d - t  + E  t  Eo  E0  = Eo  d - t  + t   E  k  k    t = Eo  d - t  +   k Q  t =  d-t +  A o  k The capacitance of the capacitor on the Fig. 8.29: Dielectric slab in the capacitor. introduction of dielectric slab becomes 207  Q Q A 0 8.11 Displacement Current: C= = = V Q  d  t  d - t +  d - t +  A 0  k  k Special cases: 1. If the dielectric fills up the entire space then A k t = d C = 0 = k C0 d ∴ capacitance of a parallel plate capacitor C increases k times i.e. k = C0 2. If the capacitor is filled with n dielectric slabs Fig. 8.31: Displacement current in the space of thickness t1, t2....... tn then this arrangement is between the plates of the capacitor. equivalent to n capacitors connected in series We know that electric current in a DC as shown in Fig. 8.30. circuit constitutes a flow of free electrons. In A 0 C = a circuit as shown in Fig 8.31, a parallel plate  t1 t 2 tn   + +............. +  capacitor with a dielectric is connected across a  k1 k 2 kn  DC source. In the conducting part of the circuit free electrons are responsible for the flow of current. But in the region between the plates of the capacitor, there are no free electrons available for conduction in the dielectric. As the circuit is closed, the current flows through the circuit and grows to its maximum value (ic) in a finite time (time constant of the circuit). The conduction current, ic is found to be same everywhere in the circuit except inside the capacitor. As the current passes through the leads of the capacitor, the electric field between the plates increases and this in Fig. 8.30 : Capacitor filled with n dielectric slabs. turn causes polarisation of the dielectric. Thus, 3. If the arrangement consists of n capacitors there is a current in the dielectric due to the in parallel with plate areas A1, A2,.............. An movement of the bound charges. The current and plate separation d due to bound charges is called displacement   C = 0 A1 k 1 + A2 k 2 +.........+ An k n d  current (id) or charge- separation current. A We can now derive an expression between if A1 = A2.............. An = then ic and id. n A 0 From Eq (8.23) we can infer that the C=  k1 + k 2 +.........+ k n  charge produced on the plates of a capacitor is dn 4. If the capacitor is filled with a conducting due to the electric field E. slab (k = ∞) then q = Akε0 E  d  Differentiating the above equation, we get C =  Co ∴ C > Co d-t dq dE  Ak  0 The capacitance thus increases by a factor dt dt --- (8.24)  d  dq/dt is the conduction current (ic)in the   conducting part of the circuit. d-t 208  dq dE ic   Ak  0 Aε 0 k dt dt (ii) Capacitance C ′ = dE i dE d  c   ic (for fixed value of A) 8.85  10  4  10 4  6.7  12  dt Ak  0 dt = The rate of change of electric field (dE/dt)  3 2 10 across the capacitor is directly proportional to = 11.86 × 10-12 F the current (ic) flowing in the conducting part Example 8.18: In a capacitor of capacitance of the circuit. 20 µF, the distance between the plates is 2 The quantity on the RHS of Eq (8.24) is mm. If a dielectric slab of width 1 mm and having the dimension of electric current and is dielectric constant 2 is inserted between the caused by the displacement of bound charges plates, what is the new capacitance ? in the dielectric of the capacitor under the Solution: Given influence of the electric field. This current, C = 20 µF = 20 × 10-6 F called displacement current (id), is equivalent d = 2 mm = 2 × 10-3 m to the rate of flow of charge (dq/dt=ic) in t = 1 × 10-3 m the conducting part of the circuit. In the k= 2 Aε 0 A 0  absence of any dielectric between the plates C= and C ′ = t of the capacitor, k =1 (for air or vacuum), the d d –t   t k displacement current id = Aε0 (dE/dt). C d – t +  As a broad generalization of displacement ⇒ = k current in a circuit containing a capacitor, it C' d  3 3  3  110 can be stated that the displacement currents do  2 10   110   20  2  not remain confined to the space between the ⇒ = C ' 2 10 3 plates of a capacitor. A displacement current (id) exists at any point in space where, time- ⇒ C ′ = 26.67 µF varying electric field (E) exists (i.e. dE/dt ≠0). 8.12 Energy Stored in a Capacitor: Example 8.17 A parallel plate capacitor A capacitor is a device used to store energy. has an area of 4 cm2 and a plate separation Charging a capacitor means transferring of 2 mm electron from one plate of the capacitor to the other. Hence work will have to be done by the (i) Calculate its capacitance battery in order to remove the electrons against (ii) What is its capacitance if the space the opposing forces. These opposing forces between the plates is filled completely with arise since the electrons are being pushed to a dielectric having dielectric constant of the negative plate which repels them and constant 6.7. electrons are removed from the positive plate Solution : Given which tends to attract them. In both the cases, A = 4 cm2 = 4 × 10-4 m2 the coulombian forces oppose the transfer d = 2 mm = 2 × 10-3 m of charges from one plate to another. As the ε0 = 8.85 × 10-12 C2 / Nm2 charge on the plate increases, its opposing Aε 0 force also increases. (i) Capacitance C = d This work done is stored in the form of 12  4 electrostatic energy in the electric field between = 8.85  10  4  10 = 1.77 × 10-12 F the plates, which can later be recovered by  3 2 10 discharging the capacitor. 209  Consider a capacitor of capacitance C The potential difference between the plates being charged by a DC source of V volts as is maintained constant at 400 volt. What is shown in Fig. 8.32. the change in the energy of capacitor if the slab is removed ? Solution : Energy stored in the capacitor with air 1 1 Ea= CV2 = ×3×10 –9 × (400)2 2 2 = 24 × 10–5 J Fig. 8.32: Capacitor charged by a DC source. when the slab of dielectric constant 3 During the process of charging, let q' is introduced between the plates of the be the charge on the capacitor and V be the capacitor, the capacitance of the capacitor increases to potential difference between the plates. Hence q' C′ = kC C= V C′ = 3 × 3 × 10–9 = 9 × 10–9 F A small amount of work is done if a small Energy stored in the capacitor with the charge dq is further transferred between the dielectric (Ed) plates. 1 q' Ed = C ' V2  dW  V dq  dq 2 C 1 Ed = × 9 × 10-9 × (400)2 Total work done in transferring the charge 2 Q q' 1 Q = 72 × 10-5 J W   dw   dq   q ' dq Change in energy = Ed– Ea = (72 - 24) × 10-5 O C CO Q = 48 ×10–5 J 1   q '  2 1 Q2 There is, therefore, an increase in the     C 2  2 C energy on introducing the slab of dielectric  0 material. This work done is stored as electrical 8.13 Van de Graaff Generator: potential energy U of the capacitor. This work Van de Graaff generator is a device used done can be expressed in different forms as to develop very high potentials of the order of follows. 107 volts. The resulting large electric fields are 1 Q2 1 1 used to accelerate charged particles (electrons, U = = CV 2 = QV  Q = CV  protons, ions) to high energies needed for 2 C 2 2 experiments to probe the small scale structure Observe and discuss of matter and for various experiments in Nuclear Physics. The energy supplied by the battery is QV It was designed by Van de Graaff (1901- but energy stored in the electric field is 1967) in the year 1931. 1 1 Principle: This generator is based on QV. The rest half QV of energy is (i) the phenomenon of Corona Discharge 2 2 wasted as heat in the connecting wires and (action of sharp points), battery itself. (ii) the property that charge given to a hollow conductor is transferred to its outer surface Example 8.19: A parallel plate air capacitor and is distributed uniformly over it, has a capacitance of 3 × 10–9 Farad. A slab (iii) if a charge is continuously supplied to an of dielectric constant 3 and thickness 3 cm insulated metallic conductor, the potential completely fills the space between the plates. of the conductor goes on increasing. 210 Construction: filled with nitrogen at high pressure. A small Fig. 8.33 shows the schematic diagram of quantity of freon gas is mixed with nitrogen to Van de Graaff generator. ensure better insulation between the vessel S and its contents. A metal plate M held opposite to the brush A on the other side of the belt is connected to the vessel S, which is earthed. Working: The electric motor connected to the pulley P1 is switched on, which begins to rotate setting the conveyor belt into motion. The DC supply is then switched on. From the pointed ends of the spray brush A, positive charge is continuously sprayed on the belt B. The belt carries this charge in the upward direction, which is collected by the collector brush C and sent to the dome shaped conductor. As the dome is hollow, the charge is Fig. 8.33: Schematic diagram of van de Graff distributed over the outer surface of the dome. generator. Its potential rises to a very high value due to P1 P2 = Pulleys the continuous accumulation of charges on it. BB = Conveyer belt The potential of the electrode I also rises to A = Spray brush this high value. C = Collector brush The positive ions such as protons or D = Dome shaped hollow conductor deuterons from a small vessel (not shown in E = Evacuated accelerating tube the figure) containing ionised hydrogen or I = Ion source deuterium are then introduced in the upper part P = DC power supply of the evacuated accelerator tube. These ions, S = Steel vessel filled with nitrogen repelled by the electrode I, are accelerated in M = Earthed metal plate the downward direction due to the very high An endless conveyor belt BB made of an fall of potential along the tube, these ions insulating material such as reinforced rubber acquire very high energy. These high energy or silk, can move over two pulleys P1 and charged particles are then directed so as to P2. The belt is kept continuously moving by strike a desired target. a motor (not shown in the figure) driving the Uses: The main use of Van de Graff generator lower pulley (P1). is to produce very high energy charged particles The spray brush A, consisting of a large having energies of the order of 10 MeV. Such number of pointed wires, is connected to the high energy particles are used positive terminal of a high voltage DC power 1. to carry out the disintegration of nuclei of supply. From this brush positive charge can different elements, be sprayed on the belt which can be collected 2. to produce radioactive isotopes, by another similar brush C. This brush is 3. to study the nuclear structure, connected to a large, dome-shaped, hollow 4. to study different types of nuclear reactions, metallic conductor D, which is mounted on 5. accelerating electrons to sterilize food and insulating pillars (not shown in the figure). E to process materials. is an evacuated accelerating tube having an electrode I at its upper end, connected to the Internet my friend dome-shaped conductor. To prevent the leakage of charge from 1. https://en.m.wikipedia.org the dome, the pulley and belt arrangement, 2. hyperphyrics.phy-astr.gsu.edu the dome and a part of the evacuated tube 3. https://www.britannica.com/science are enclosed inside a large steel vessel S, 4. https://www.khanacademy.org>in-i 211  Exercises Q1. Choose the correct option i) A parallel plate capacitor is charged and qQ qQ (C) (D) then isolated. The effect of increasing 6 0 L 4 0 L the plate separation on charge, potential, v) A parallel plate capacitor has circular capacitance respectively are plates of radius 8 cm and plate separation (A) Constant, decreases, decreases 1mm. What will be the charge on the (B) Increases, decreases, decreases plates if a potential difference of 100 V (C) Constant, decreases, increases is applied? (D) Constant, increases, decreases (A) 1.78 × 10-8 C (B) 1.78 × 10-5 C ii) A slab of material of dielectric constant (C) 4.3 × 104 C (D) 2 × 10-9 C k has the same area A as the plates of a Q2. Answer in brief. parallel plate capacitor and has thickness i) A charge q is moved from a point A (3/4d), where d is the separation of the above a dipole of dipole moment p to plates. The change in capacitance when a point B below the dipole in equitorial the slab is inserted between the plates is plane without acceleration. Find the A 0  k  3  (A) C  work done in this process. d  4 k  A (B) C  0  2 k  d  k 3 A 0  k  3  (C) C  d  2 k  A  4 k  (D) C  0  d  k  3  iii) Energy stored in a capacitor and ii) If the difference between the radii of the dissipated during charging a capacitor two spheres of a spherical capacitor is bear a ratio. increased, state whether the capacitance (A) 1:1 (B) 1:2 will increase or decrease. (C) 2:1 (D) 1:3 iii) A metal plate is introduced between iv) Charge +q and -q are placed at points the plates of a charged parallel plate A and B respectively which are distance capacitor. What is its effect on the 2L apart. C is the mid point of A and B. capacitance of the capacitor? The work done in moving a charge +Q iv) The safest way to protect yourself from along the semicircle CRD as shown in lightening is to be inside a car. Justify. the figure below is v) A spherical shell of radius b with charge Q is expanded to a radius a. Find the work done by the electrical forces in the process. 3. A dipole with its charges, -q and +q located at the points (0, -b, 0) and (0 +b, 0) is present in a uniform electric field E qQ qQ whose equipotential surfaces are planes (A) (B) parallel to the YZ planes. 6 0 L 2 0 L 212  (a) What is the direction of the electric in which all the dipoles are perpendicular field E? (b) How much torque would to the field, θ2 = 90°.[Ans: 1.575 × 10-3 J] the dipole experience in this field? [(a) 11. A charge 6 µC is placed at the origin along/parallel to X axis, (b) τ = 2bqE] and another charge –5 µC is placed on 4. Three charges -q, +Q and -q are placed the y axis at a position A (0, 6.0) m. at equal distance on straight line. If the potential energy of the system of the three charges is zero, then what is the ratio of Q:q? [(Q : q = 1 : 4)] 5. A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, a) Calculate the total electric potential the energy stored in it, the electric field, at the point P whose coordinates are charge stored and voltage will increase, (8.0, 0) m decrease or remain constant. b) Calculate the work done to bring a 6. Find the ratio of the potential differences proton from infinity to the point P. that must be applied across the parallel and series combination of two capacitors What is the significance of the sign of C1 and C2 with their capacitances in the the work done ? ratio 1:2, so that the energy stored in [Ans: (a) Vp = 2.25 × 103 V these two cases becomes the same. (b) W = 3.6 × 10-16 J] [Vp : Vs = √ 2:3] 12. In a parallel plate capacitor with air 7. Two charges of magnitudes -4Q and between the plates, each plate has an +2Q are located at points (2a, 0) and (5a, area of 6 × 10–3 m2 and the separation 0) respectively. What is the electric flux between the plates is 2 mm. a) Calculate due to these charges through a sphere of the capacitance of the capacitor, b) If this radius 4a with its centre at the origin? capacitor is connected to 100 V supply, 8. A 6 µF capacitor is charged by a 300 what would be the charge on each plate? V supply. It is then disconnected c) How would charge on the plates be from the supply and is connected to affected if a 2 mm thick mica sheet of another uncharged 3µF capacitor. How k = 6 is inserted between the plates while much electrostatic energy of the first capacitor is lost in the form of heat and the voltage supply remains connected ? electromagnetic radiation ? [Ans: (a) 2.655 × 10-11 F, [Ans: 9 × 10-2 J] (b) 2.655 × 10-9 C, (c) 15.93 × 10-9 C] 9. One hundred twenty five small liquid 13. Find the equivalent capacitance between drops, each carrying a charge of P and Q. Given, area of each plate = A 0.5 µC and each of diameter 0.1 m form and separation between plates = d. a bigger drop. Calculate the potential at 2 Aε 0 4 Aε 0 [Ans: (a) (b) ] the surface of the bigger drop. d d [Ans: 2.25 × 106 V] 10. The dipole moment of a water molecule is 6.3 × 10–30 Cm. A sample of water contains 1021 molecules, whose dipole moments are all oriented in an electric field of strength 2.5 × 105 N /C. Calculate the work to be done to rotate the dipoles from their initial orientation θ1 = 0 to one 213