JEE Advanced 2022 · Paper 1 · Physics Q13

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PhysicsMultiple correct+4 / -2Question 31 of 54

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 regular hexagon of side a with centre O. Six point charges lie outside it, one beyond the midpoint of each side, on the perpendicular from O through that midpoint; each side bisects the perpendicular from its charge (90° mark at the top side). Five charges are q; the one below the bottom side is x.

When

x = q

, the magnitude of the electric field at O is zero.

When

x = -q

, the magnitude of the electric field at O is

\dfrac{q}{6\pi\epsilon_0a^2}

When

x = 2q

, the potential at O is

\dfrac{7q}{4\sqrt3\pi\epsilon_0a}

When

x = -3q

, the potential at O is

-\dfrac{3q}{4\sqrt3\pi\epsilon_0a}

Answer and worked solution

Official answer: A, B, C

Distances. Each charge is at twice the apothem from O:
$r=2\cdot\frac{\sqrt3}2a=\sqrt3\,a.$

(A) True. With six equal charges placed symmetrically, the field at O cancels.

(B) True. With $x=-q$ , the field equals that of six $q$ charges (zero) plus a charge $-2q$ at the position of $x$ :
$E=\frac{1}{4\pi\epsilon_0}\frac{2q}{3a^2}=\frac{q}{6\pi\epsilon_0a^2}.$

(C) True. With $x=2q$ :
$V=\frac{7q}{4\pi\epsilon_0\sqrt3\,a}.$

(D) False. With $x=-3q$ the total charge is $2q$ , so $V=+\frac{2q}{4\sqrt3\pi\epsilon_0a}$ .

Topics:Electric field Electric potential Coulomb's law and superposition

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JEE Advanced 2022 · Paper 1, Physics question 13