Page 9 - English Class X_cbse new (FINAL).cdr

S. Solution Marking S. Solution Marking
No. Scheme No. Scheme

12. Here r1 = 1m, r2 = 3m, r3 = 9m .... [1] 15. Force on charge (+q) = + q E along the direction E [1]
[1] Cont. Force on charge (-q) = -q E along the opp. Direction of E
Electric field, E = q  1 + 1 + 1  [2]
4π ∈0  + ..... The net force on the dipole is zero since E is uniform.
 (1)2 (3)2 (9)2


Applying G.P. (sum of series) These force being equal, unlike and parallel form a couple, which
rotates the dipole in clock-wise direction.

1 ∴ Magnitude of torque = force × arm of couple
= q   = q × 9 = 1 ⋅ 9q NC−1
4π ∈0  8  4π ∈0 8 4π ∈0 8 τ = F.AC = q E A B sin ¸ =(q E) 2a sin θ

9 τ =q(2a) E sin θ

13. Electric dipole moment of a dipole: The magnitude of electric τ = PE sin θ [∵P = q(2a)]
dipole moment of a dipole is equal to product of magnitude of either
charge and the distance between them. ∴ τ =P×E [1]

( )p = q 2a or p = q(2a) or P = q(2a) The direction of τ is given by right hand screw rule and is normal to [3]
P and E .
The SI unit of dipole moment is Coulomb-metre (C-m) [1]

16.

14. The dipole is in stable equilibrium when electric dipole is in the direc- [1]
tion of electric field.

15. (a) Dipole in a Uniform External Field: Let P be an axial point at distance r from the centre of the dipole. [1]

+q [1] Electric potential at point P will be [1]
A [2]
V=V1 + V2 = − 1 r q + 1 ⋅ q
θ 4π ∈0 +a 4π ∈0 r −a 7

= q  1 − r 1 = q ⋅ 2a
4π ∈0  r −a + a  4π ∈0 r2 − a2

Consider an electric dipole consisting of two equal and opposite point = 1 ⋅ r2 p [∵ p = q (2a)]
charges separated by a small distance AB = 2a having dipole mo- 4π ∈0 − a2
ment
For a far away point, r >> a
P = q(2a)
∴ V= 1 ⋅ p
Let the dipole held in a uniform external electric field E at an angle . 4π ∈0 r2