Page 6 - English Class X_cbse new (FINAL).cdr

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

2. At large distances (a >> ), this reduces to [1] 2. we know [1]
cont. cont. τ = PE sin θ [3]

E = − 2q p ...(iv) here θ = 180°
4πε0a3 τ = PE sin 180°
τ =0
∵ p = q × 2ap

∴E = −p (a >> ) but Net force on dipole -2qE hence dipole is in state of unstable
4πε0a3 equilibrium.

(ii) Depicting the orientation of the dipole in (i) stable equilibrium in a 3. Suppose we have a ring of radius a that carries uniformly distributed
uniform electric field. positive charge q.

E dl dE sin θ

r

P dE

-q +q a θ θ dE cos θ
xθ θ

ra dE

dl dE sin θ

We know As the total charge q is uniformly distributed, the charge dq on the
τ = PE sin θ element dl is
when dipole is parallel to electric field θ = 0
τ = PE sin 0° dq = q ⋅ dl
τ = 0, hence dipole is in permanent equilibrium. 2πa
(ii) unstable equilibrium in a uniform electric field.
∴ T he magnitude of the electric field produced by the element dl at
E
the axial point P is
+q -q
dE = k⋅ dq = kq ⋅ dl
r2 2πa r2

The electric field dE has two components.

(i) The axial component dEcosθ and

(ii) The perpendicular component dEsin θ.

Since the perpendicular component of any two diametrically opposite
elements are equal and opposite they cancel out in pairs. Only the
axial components will add up to produce the resultant field.

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