Electrostat-02-OBJECTIVE SOLVED

SOLVED PROBLEMS (OBJECTIVE) 1. Two equal point charges are fixed at x = - a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to : (a) x (b) x2 (c) x3 (d) 1/x. Ans. (b) 2kQ Solution : Ui = a q kqQ kqQ a a q Q q Uf = a  x  a  x a+x (a-x) q Q q 2kqQa = a 2  x 2  a  1  U – U = 2kqQ  a 2 (1  x 2 / a 2 )   a  f i   2kqQ  x 2      U  a 1  a2  1    2kqQ  x 2    a a 2   U  x2.   2. Three charges Q, +q and +q are placed at the vertices of a right Q angled isosceles triangle as shown in the figure. The net electro- static energy of the configuration is zero., if Q is equal to :  q  2q (a) 1  2 (b) 2  2 (c) -2q (d) +q . Ans. (b) Solution : Net electrostatic energy U  kQq   kqq a a +q +q For U = 0; Qq  a  qq = 0   Qq    q 2   2 a  a  2  1   q  Q     2 a  a  2   Q = -q    2  1  2  = -q   2  2   3. Two point charges +q and –q are held fixed at (-d, 0) and (d, 0) respectively of a (X, Y) coordinate system. Then → (a) T→he electric field E at all points on the X-axis has the same direction. (b) E at all points on the Y-axis is along iˆ . (c) Work has to be done in bringing a test charge from infinity to the origin. (d) The dipole moment is 2qd directed along iˆ . Ans. (b) Solution : The diagrammatic representation of the given problem is shown in fig. The electrical field same direction. The electrical field → at all points on the X-axis will not have the → at all points on the Y-axis will be parallel to E E  Eˆi E the X-axis (i.e. iˆ direction). The electric potential at the origin due to both the charge is zero, hence, no work is done in bringing a test charge from infinity to the origin. +q -q X Dipole moment is directed from the –q charge to the +q charge (i.e. –x direction). (-d, 0) O (d, 0) 4. A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0, … inf. on the x-axis and a charge –q is fixed at each of the points x = 2x0, x = 4x0, x = 6x0, …. inf. Here x0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/ 4πε 0 r. Then, the potential at the origin due to the above system of charges is : q (a) 0 (b) (c)  (d) Ans. (d) Solution: Potential at origin 8πε 0 x 0 In2 q In2 4πε 0 x 0 . q 1 1  1  1  1  1   V = 4 q  0 x0  ln 2 2 3 4 ........ 5  = 4 x 0 5 Two identical thin rings, each of radius R, are coaxially placed a distance R apart. If Q1 and Q2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring that of the other is : (a) zero (b) q (Q1 – Q2) (c) q 2 Q1  Q 2 2  1 (d) q (Q /Q ) . Ans. (b) 4πε0 R 1 2 4πε 0 R Solution: The potential at A due to the charge on the ring 1 is given as : VA1 = Q1 4πε 0 . 1 R The potential at A due to the charge Q2 on the ring 2 is given as : VA2 = Q2 40 . 1 1 2 Q2 . = 4πε0 1  R 2  R 2 1 . Q 2 . 1 4πε 0 R Total potential at A is 1  Q 2  V = V + V = 4πε R Q1  2  A A1 A2 0   The potential energy of charge q at A is q  Q 2  U = V = 4πε R Q1  2  A Aq 0   Similarly, the potential energy of charge q at B is UB = q 4πε  Q1  Q  R 0   The work done in moving a charge q from point A to B is : W =  U = – U + U B  q . Q1  Q  Q  Q2  4 R  2 1 2  0   q Q  Q  1   = 4 R 2 1     6. There are points on a straight line jointing two fixed opposite charges. There is : (a) no point where potential is zero (b) only one point where potential is zero (c) no point where electric field is zero (d) only one point where electric field is zero. Ans. (b) Solution: Let two opposite charges +q and –q be situated at points A and B respectively. E1  1 q 4πε 0 a 2 E 2  1 4πε 0 q d  a2 A +q E2 j q B E = E1 + E2  q  1  1  4πε a 2 d  a2   q 4πε 0 d 2  2ad  a 2  a 2  a 2 d  a2 Hence, there can be more that one point where electric field is zero. V1 = 1 4πε 0 a ; V2  1 4πε0  q d  a V = V1 + V2 = q  1  1   qd  2a 4πε 0  a d  a 4πε 0 ad  a  Potential is zero only at d=2a or a=d/2. 7. A certain charge Q is divided into two parts q and (Q-q). For the maximum coulomb force between them, the ratio (q/Q) is : (a) 1/16 (b) 1/8 (c) 1/4 (d) 1/2. Ans. (d) Solution: 1 qQ  q F 4πε 0 r 2 for F to be maximum, dF  0 dq 1 4πε 0 . 1 Q  q  q 1 0 r 2 Q - 2q = 0 q  1 . Q 2 8. A charge is situated at a certain distance from an electric dipole in the end-on position experiences a force F. if the distance of the charge is doubled, the force acting on the charge will be : (a) F/4 (b) F/8 (c) 2F (d) F/2. Ans. (b) Solution: E  1 2p 4πε0 r 3 E  1 r 3 F  1 r 3 Hence, the force will become F/8. 9. A parallel plate capacitor of area A, plate separation d and capaci- tance C is filled with three different dielectric materials having di- electric constants k1, k2 and k3 as shown. If a single dielectric material is to be used to have the same capacitance C in this ca- pacitor, then its dielectric constant k is given by : A = area of plates (a) 1  1  1  1 (b) 1  1  1 k k1 k 2 2k 3 k k1  k 2 2k 3 (c) Ans. (b) k  k1k 2 k1  k 2  2k 3 (d) k = k1 + k2 + 2k . Solution : The given circuit is equivalent to the following circuit C  A/ 2k10 ; 1 d / 2 C  A / 2k20 ; 2 d / 2 C  A / 2 k30 3 d / 2 i.e. C  Ak10 ; 1 d C  Ak20 ; C 2 d 3  Ak30 d The capacitors C1 and C2 are in parallel, their resultant is C12 = C1 + C2 = A0 k d 1  k2  The capacitors C12 and C3 are in series and their resultant is 1  1  1 = d  d C C12 C3 Ak1  k2 0 2Ak30 or C  Ak0 where 1  1  1 d k k1  k 2 2k 3 10. Two identical metal plates are given positive charges Q1 and Q2 ( VB (c) VA < VC (d) VA > VC . 1 C Ans. (b) Solution: Direction of electric field is in the direction of potential drop E  VA  VB VA  VC B A(0, 0) 1 17. An electron of mass me initially at rest moves through a certain distance in a uniform electric field in time t1. A proton of mass mp also initially at rest takes time t2 to move through an equal distance in this t2 uniform electric field. Neglecting the effect of gravity the ratio of t1 is nearly equal to : 1  mp  2 (a) 1 (b)    me  1  me  2 (c)   (d) 1836.  mp    Ans. (b) Solution : Force on a charge particle in a uniform electric field F = q E The acceleration imparted to the particle is a  qE m The distance traveled by the particle in time t is d  1 at 2  1  qE t 2   2 2  m  For the given problem p  t e mp me p  mp 2 me t p  t e 18. The arc AB with the centre C and the infinitely long wire having linear charge density  are lying in the same plane. The minimum amount of work to be expended to move a point charge q0 from point A to B through a circular path AB of radius a is equal to : (a) q0 ln 2 (b) q0 ln 3 20 3 20 2 (c) q0 ln 2 (d) q0 . Ans. (b) Solution: 20 3 E   20 x VB  2a dx  dV  E dx   2 A  0 3a  VB  VA   2  3    2 0   work done by agent = q 0  20 In 3 2 19. A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and work done on the system, in question, in the process of inserting the slab, then : (a) Q  ε 0 AV d V (b) Q  ε 0 KAV d ε 0 AV 2  1  (c) E  kd (d) W  2d 1  K  . Ans. (a), (c), (d) Solution : New capacitance C  K0A d  KC   ε A C     Voltage V  Q  Q  V C KC K Electric field E  V  V d Kd Work done w  U  1 CV2  1 CV2 2 2  1  V 2 2  1 ε 0A V 2 ε 0 A 2  w  U 2 KC K 2  CV   2  d K  d V       ε 0 AV 2  1  w 2d 1  K   