JEE Advanced 2021 Paper 1 · Q09 · Electric Potential

Question

Two point charges $-Q$ and $+Q/\sqrt{3}$ are placed in the $xy$-plane at the origin $(0,\,0)$ and at the point $(2,\,0)$, respectively. All lengths are measured in metres. This arrangement produces an equipotential circle of radius $R$ and potential $V = 0$ in the $xy$-plane, with its centre at $(b,\,0)$. The value of $R$ is ___ metre.

SolveKar AI · Accepted Answer

The locus of V = 0 for two unequal point charges is the Apollonius circle. Let the circle cross the x-axis at (a, 0) and (C, 0). Setting V = 0: k(−Q)/r₁ + k(Q/√3)/r₂ = 0 ⇒ r₁/r₂ = √3. Solving |x|/|x − 2| = √3 gives x = 2√3/(√3+1) and x = 2√3/(√3−1). Diameter = (C − a)/2: C − a = 2√3/(√3−1) − 2√3/(√3+1) = 2√3·[2/(3−1)] = 2√3. Radius R = (C − a)/2 = √3 ≈ 1.73 m.

JEE Advanced 2021 Paper 1 · Q09 · Electric Potential

Solution