JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Electrostatics

Figure shows a rod AB, which is bent in a

12 0^{\circ}

circular arc of radius

R

. A charge (

- Q

) is uniformly distributed over rod

A B

. What is the electric field

E

at the centre of curvature O ?

Answer: (B)

Solution

The arc carrying

- Q

subtends

12 0^{\circ}

symmetrically about the

x

-axis, so transverse contributions cancel and the net field at the centre lies along the axis. For an arc of half-angle

α

and linear charge density

λ

, the axial field is

E = \frac{2 k λ sin α}{R}

with

k = \frac{1}{4 π ε _{0}}

. Here

α = 6 0^{\circ}

and

λ = \frac{Q}{R ( 2 π /3 )} = \frac{3 Q}{2 π R}

, so

E = \frac{1}{4 π ε _{0}} \cdot \frac{2 sin 6 0 ^{\circ}}{R} \cdot \frac{3 Q}{2 π R} = \frac{3 3 Q}{8 π ^{2} ε _{0} R ^{2}} .

Since the charge is negative the field at

O

points toward the arc, along

+ \hat{i}

E = \frac{3 3 Q}{8 π ^{2} ε _{0} R ^{2}} \hat{i}

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option