JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Electrostatics

Six charges are placed around a regular hexagon of side length a as shown in the figure. Five of them have charge

q

, and the remaining one has charge

x

. The perpendicular from each charge to the nearest hexagon side passes through the center O of the hexagon and is bisected by the side.

Which of the following statement(s) is(are) correct in SI units?

(A)When $x = q$ , the magnitude of the electric field at O is zero.✓ Correct
(B)When $x = - q$ , the magnitude of the electric field at O is $\frac{q}{6 π ϵ _{0} a ^{2}}$ .✓ Correct
(C)When $x = 2 q$ , the potential at O is $\frac{7 q}{4 3 π ϵ _{0} a}$ .✓ Correct
(D)When $x = - 3 q$ , the potential at O is $\frac{3 q}{4 3 π ϵ _{0} a}$ .

Answer: (A), (B), (C)

Solution

Distance from each charge to center:

d = \frac{a}{3}

(A) When

x = q

, by symmetry all six charges produce zero net field at O. Correct.

(B) When

x = - q

, the net field is due to two charges

2 q

(the difference):

E = \frac{2 k q}{( 2 d ) ^{2}} = \frac{2 q}{4 π ϵ _{0} \times 4 \times 3 a ^{2} /3} = \frac{q}{6 π ϵ _{0} a ^{2}}

Correct.

(C) When

x = 2 q

, total charge is

7 q

, potential:

V = \frac{7 k q}{d} = \frac{7 q}{4 π ϵ _{0} \times a / 3} = \frac{7 q 3}{4 π ϵ _{0} a} = \frac{7 q}{4 3 π ϵ _{0} a}

Correct.

(D) When

x = - 3 q

, total charge is

2 q

, potential:

V = \frac{2 k q}{d} = \frac{2 q}{4 3 π ϵ _{0} a}

This does not match option (D). Incorrect.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · One or more correct options

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Electrostatics

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