JEE Advanced 2026 · Paper 2 · Physics Q7

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PhysicsMultiple correct+4 / -1Question 25 of 54

Two charges $Q_1=q$ and $Q_2=mq$ are placed at the points $P_1(a,b)$ and $P_2(ma,mb)$ , respectively, in the $XY$ plane, where $a,b\ne0$ and $m\ne0,1$ . If $V_1$ is the potential at a point in the $XY$ plane due to charge $Q_1$ and $V_2$ is the potential at that point due to charge $Q_2$ . Correct statement(s) for the points at which $|V_1|=|V_2|$ is/are:

For

m=-1

, locus of these points is

ax+by=0

For

m=2

, the locus of these points is a circle of radius

\frac23\sqrt{a^2+b^2}

centered at

\left(\frac23a,\frac23b\right)

For

m=-2

, the locus of these points is a circle of radius

2\sqrt{a^2+b^2}

centered at

(2a,2b)

For

m=-3

, locus of these points is

3bx+3ay=0

Answer and worked solution

Official answer: A, B, C

Write $\vec p=(a,b)$ . Then $P_1=\vec p$ , $P_2=m\vec p$ , and $|V_1|=|V_2|$ means $|\vec X-m\vec p|=|m|\,|\vec X-\vec p|$ .

(A) $m=-1$ : $|\vec X+\vec p|=|\vec X-\vec p|$ gives $\vec X\cdot\vec p=0$ , i.e. $ax+by=0$ . True.
(B) $m=2$ : $|\vec X-2\vec p|^2=4|\vec X-\vec p|^2$ gives $3|\vec X|^2-4\vec X\cdot\vec p=0$ , i.e. $\left|\vec X-\tfrac23\vec p\right|=\tfrac23|\vec p|$ . A circle with centre $\left(\frac23a,\frac23b\right)$ and radius $\frac23\sqrt{a^2+b^2}$ . True.
(C) $m=-2$ : $|\vec X+2\vec p|^2=4|\vec X-\vec p|^2$ gives $3|\vec X|^2-12\vec X\cdot\vec p=0$ , i.e. $|\vec X-2\vec p|=2|\vec p|$ . Centre $(2a,2b)$ , radius $2\sqrt{a^2+b^2}$ . True.
(D) $m=-3$ : $|m|\ne1$ gives an Apollonius circle, not a straight line. False.

Topics:Electric potential Equipotential surfaces

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JEE Advanced 2026 · Paper 2, Physics question 7