JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Electrostatics

Find the electric field at point

P

(as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length

L

carrying a charge

Q

. The distance of the point

P

from the centre of the rod is

a = \frac{3}{2} L

(A) $\frac{Q}{3 π ε _{0} L ^{2}}$
(B) $\frac{Q}{2 3 π ε _{0} L ^{2}}$ ✓ Correct
(C) $\frac{Q}{4 π ε _{0} L ^{2}}$
(D) $\frac{3 Q}{4 π ε _{0} L ^{2}}$

Answer: (B)

Solution

The rod carries linear charge density

λ = \frac{Q}{L}

. At a point on the perpendicular bisector the components parallel to the rod cancel in pairs, and the resultant field points away from the rod with magnitude

E = \frac{λ}{4 π ε _{0} a} (2 sin θ), sin θ = \frac{L /2}{a ^{2} + L ^{2} /4}

where

θ

is the half angle the rod subtends at the point. With

a = \frac{3}{2} L

a^{2} + \frac{L ^{2}}{4} = \frac{3 L ^{2}}{4} + \frac{L ^{2}}{4} = L^{2}

, so

sin θ = \frac{L /2}{L} = \frac{1}{2}

and the bracket is

1

. Hence

E = \frac{λ}{4 π ε _{0} a} = \frac{Q / L}{4 π ε _{0} ( \frac{3}{2} L )} = \frac{Q}{2 3 π ε _{0} L ^{2}}

so the field at the point is

\frac{Q}{2 3 π ε _{0} L ^{2}}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Electrostatics

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