JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Electrostatics

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Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge

(+ q)

each, while

2, 4, 6, 8, 10

have charge

(- q)

each. The potential

V

and the electric field

E

at the centre of the circle are respectively:
(Take

V = 0

at infinity)

(A) $V = \frac{10 q}{4 π ϵ _{0} R}; E = \frac{10 q}{4 π ϵ _{0} R ^{2}}$
(B) $V = 0; E = \frac{10 q}{4 π ϵ _{0} R ^{2}}$
(C) $V = 0; E = 0$ ✓ Correct
(D) $V = \frac{10 q}{4 π ϵ _{0} R}; E = 0$

Answer: (C)

Solution

Potential is a scalar, so the ten equal charges simply add algebraically at the centre. With five charges of

+ q

and five of

- q

, each a distance

R

away,

V = \frac{1}{4 π ϵ _{0} R} (5 q - 5 q) = 0

. The field, however, is a vector. The charges sit at equal angular steps of

3 6^{\circ}

with their signs alternating around the ring, and when the field contributions of all ten are drawn at the centre they fall into a symmetric set of equally spaced directions that cancel in pairs and overall. Hence both the potential and the electric field at the centre are zero.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Electrostatics

Solution

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