Figure shows three spherical and equipotential surfaces 1,2 and 3 round a point charge q. The potential difference V₁–V₂ = V₂ – V₃. If t₁ and t₂ be the distance between them. Then

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$t_{1} = t_{2}$

$t_{1} < t_{2}$

$t_{1} \leq t_{2}$

$t_{1} > t_{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} V_{1} - V_{2} = kq (\frac{1}{r_{1}} - \frac{1}{r_{2}}) \\ r_{2} - r_{1} = \frac{(V_{1} - V_{2}) r_{1} r_{2}}{kq}; but (r_{2} - r_{1}) = t \\ ∴ t α r_{1} r_{2} \end{array}$

$if P.D is constant then (r_{2} - r_{1}) = t$

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