Electric Charges and Fields

1.14 Since the charges 1 & 2 are attracted towards + ve, their charges will be – ve. The charge 3 is attracted towards – ve, hence its charge will be +ve.

The charge to mass ratio (emf) is directly proportional to the displacement, charge 3 will have the highest charge to mass ratio.

1.13 When the third uncharged sphere C is brought in contact with the sphere A, then the charge is shared and becomes half. Then

 $q_A$ = $\frac{q}{2}$and  = $q_C$= $\frac{q}{2}$

When the charged sphere C is brought in contact with charged sphere B, the charge between both the sphere is shared and becomes half

 $q_B,q_C=\frac{1}{2}$ (q + = $\frac{q}{2})$$\frac{3q}{4}$ 

Hence the force of repulsion between sphere A and B can be given as

F = 
$\frac{1}{4\pi {?}_0}$ $*\frac{q_1{q}_2}{r^2}$, where ${?}_0$  = Permittivity of free space = 8.854  $*{10}^{-12}$ $C^2{N}^{-\mathrm{1 \; \; }}$$m^{-2}$ = =$\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}}$ = $*\frac{(\frac{q}{2}*\frac{3q}{4})}{{0.5}^2}$ =$\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}}$$*\frac{3*q^2}{{8*0.5}^2}$ 
= $\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}}$ $*\frac{3*{(6.5\mathrm{}*{10}^{-7})}^2}{{8*0.5}^2}$  = 5.695$*{10}^{-3}$N

1.12 (a) Charge on sphere A, $q_A$ = 6.5 $*{10}^{-7}$ C

Charge on sphere B, $q_B$ = 6.5$*{10}^{-7}$ C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between two spheres

F = $\frac{1}{4\pi {?}_0}$ $*\frac{q_1{q}_2}{r^2}$ , where ${?}_0$= Permittivity of free space = 8.854$*{10}^{-12}$ 
  $C^2{N}^{-1}$$m^{-2}$Therefore, F =  $\frac{1}{4*\pi *{8.854\mathrm{}*{10}^{-12}}_{}}$ $*\frac{6.5\mathrm{}*{10}^{-7}*6.5\mathrm{}*{10}^{-7}}{{0.5}^2}$N = 0.0152 N = 1.52  $*{10}^{-2}$N

(b) Charge on sphere A, $q_A$ = 2 $*$6.5  $*{10}^{-7}$C = 1.3

Charge on sphere B, $q_B$ = 2 $*$ 6.5 $*{10}^{-7}$C = 1.3

Distance between the spheres, r =
$\frac{50}{2}$ = 25 cm = 0.25 m

Force of repulsion between two spheres

F =  $\frac{1}{4\pi {?}_0}$ $*\frac{q_1{q}_2}{r^2}$, where ${?}_0$ = Permittivity of free space = 8.854  $*{10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$Therefore, F = 
 $\frac{1}{4*\pi *{8.854\mathrm{}*{10}^{-12}}_{}}$ $*\frac{1.3\mathrm{}*{10}^{-6}*1.3\mathrm{}*{10}^{-6}}{{0.25}^2}$N = 0.243 N

Ans.1.12

(a) Charge on sphere A, $q_A$ = 6.5 $*{10}^{-7}$ C

Charge on sphere B, $q_B$ = 6.5$*{10}^{-7}$ C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between two spheres

F = $\frac{1}{4?{?}_0}$ $*\frac{q_1{q}_2}{r^2}$ , where ${?}_0$= Permittivity of free space = 8.854$*{10}^{-12}$ 
  $C^2{N}^{-1}$$m^{-2}$Therefore, F =  $\frac{1}{4*?*{8.854\mathrm{}*{10}^{-12}}_{}}$ $*\frac{6.5\mathrm{}*{10}^{-7}*6.5\mathrm{}*{10}^{-7}}{{0.5}^2}$N = 0.0152 N = 1.52  $*{10}^{-2}$N

(b) Charge on sphere A, $q_A$ = 2 $*$6.5  $*{10}^{-7}$C = 1.3

Charge on sphere B, $q_B$ = 2 $*$ 6.5 $*{10}^{-7}$C = 1.3

Distance between the spheres, r = $$ = 25 cm = 0.25 m

Force of repulsion between two spheres

F =  $\frac{1}{4?{?}_0}$ $*\frac{q_1{q}_2}{r^2}$, where ${?}_0$ = Permittivity of free space = 8.854  $*{10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$Therefore, F = 
 $\frac{1}{4*?*{8.854\mathrm{}*{10}^{-12}}_{}}$ $*\frac{1.3\mathrm{}*{10}^{-6}*1.3\mathrm{}*{10}^{-6}}{{0.25}^2}$N = 0.243 N

1.11

(a) When polythene rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene negatively charged.

Amount of charge on the polythene piece, q = -3 $*{10}^{-7}$ C.

Amount of charge of 1 electron e = -1.6 $*10-19$

So number of electron transferred from wool to polythene

=
-$\frac{3\mathrm{}*{10}^{-7}}{-1.6\mathrm{}*{10}^{-19}}$1.875$*{10}^{12}$

(b) Since electron has a mass, so there will be transfer of mass also.

Mass of single electron,  $<math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></math>$ = 9.1 $*{10}^{-31}$ kg

Total mass transferred from wool to polythene = 1.875 $*{10}^{12}*$9.1 $*{10}^{-31}$ kg

= 1.706 $*{10}^{-18}$kg$?$ negligible

Hence a negligible amount of mass is transferred from wool to polythene.

Ans.1.10 Electric dipole moment, p = 4 $\mathit{10}-9$
 C m

Angle made by p with a uniform electric field,  $?$ = 30 $°$

Electric field, E = 5  $*{10}^4$N$C^{-1}$

Torque acting on the dipole is given by $?$= pE $\sin ??$ = 4 $*{10}^{-9}*$ 5 $*{10}^4*\sin ?30°$

= 1 $*{10}^{-4}$ Nm

Therefore, the magnitude of the torque acting on the dipole is ${10}^{-4}$ Nm

1.9

At A, the amount of charge, $q_A$2.5 $*{10}^{-7}$ C

At B, the amount of charge, $q_B$
 = –2.5 $*{10}^{-7}$ C

Total charge of the system, q = $q_A$
+ $q_B$ = 0

Distance between two charges at point A and B = 15 + 15 = 30 cm = 0.3 m

Electric dipole moment of the system is given by p  = $q_A*d$ =  $q_B*d$ = 2.5$*{10}^{-7}*0.3$= 7.5$*{10}^{-8}$  C m, along positive z axis.

1.7

(a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to another.

(b) The electric field intensity will show two directions at that point where two filed lines crosses. This is not possible. Hence they do not cross.