Electric Charges and Fields

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

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

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

Force of repulsion between two spheres

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

(b) Cha

Ans.1.12

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

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

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

Force of repulsion between two spheres

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

(b) Charge on

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 e

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, alon

1.8 (a) The situation is represented in the following figure.

O is the midpoint of AB, hence OA = OB = 10 cm = 10$\times {10}^{-2}$ $m^2$

The electric field at O caused by the charge at A is given by

 $E_1$ =$\left|\frac{4\pi {?}_0}{}\mathrm{}\times \frac{q_A}{r^2}\right|$ along OB and the electric field at O caused by the charge at B is given by,

 $E_2$ =$\left|\frac{1}{4\pi {?}_0}\mathrm{}\times \frac{q_B}{r^2}\right|$along OB

where ${?}_0$= Pe

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 intensi

1.6

In the adjoining figure ABCD is a square with sides AB = BC = CD = DA = 10 cm

Diagonals, AC = BD  = $\sqrt[]{{10}^2+{10}^2}$ = $10\sqrt[]{2}$cm

AO = OC = DO = BO = $5\sqrt[]{2}$ cm

At the centre of the square ABCD, O, a charge of 1 $\mu Cisplaced.$$\text{exception 25:}$

The force of repulsion between the charges placed at A and at O is equal in magnitude but opposite in

Ans.1.5 When two bodies are rubbed against each other, it produces charges of equal magnitude in both the bodies but of opposite in nature. Hence the net charges of the two bodies are zero. When a glass rod is rubbed with a silk cloth, similar phenomena occur. This is as per the law of conservation