JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Electrostatics

A uniformly charged disc of radius R having surface charge density

σ

is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin :-

(A) $E = \frac{σ}{2 ε _{0}} (1 + \frac{Z}{( Z ^{2} + R ^{2} ) ^{1/2}})$
(B) $E = \frac{σ}{2 ε _{0}} (1 - \frac{Z}{( Z ^{2} + R ^{2} ) ^{1/2}})$ ✓ Correct
(C) $E = \frac{2 ε _{0}}{σ} (\frac{1}{( Z ^{2} + R ^{2} ) ^{1/2}} + Z)$
(D) $E = \frac{σ}{2 ε _{0}} (\frac{1}{( Z ^{2} + R ^{2} )} + \frac{1}{Z ^{2}})$

Answer: (B)

Solution

Treating the disc as concentric rings of radius

r

and width

d r

, each ring of charge

σ 2 π r d r

contributes an axial field

d E = \frac{σ 2 π r d r}{4 π ε _{0}} \frac{Z}{( Z ^{2} + r ^{2} ) ^{3/2}}

. Integrating from

r = 0

R

gives

E = \frac{σ}{2 ε _{0}} (1 - \frac{Z}{Z ^{2} + R ^{2}}) .

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option