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Q.

Figure shows three spherical and equipotential surfaces A B, and C around a point charge q. The potential $V_{A} - V_{B} = V_{B} - V_{C} .$ If $t_{1}$ and $t_{2}$ be the distances between them, then

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a

$t_{1} = t_{1}$

b

$t_{1} > t_{2}$

c

$t_{1} < t_{2}$

d

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

answer is C.

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Detailed Solution

Potential difference between two equipotential surfaces A and B.
$V_{A} - V_{B} = k q (\frac{1}{r_{A}} - \frac{1}{r_{B}}) = k q (\frac{r_{B} - r_{A}}{r_{A} r_{B}}) = \frac{k q t_{1}}{r_{A} r_{B}} t_{1} = \frac{(V_{A} - V_{B}) r_{A} r_{B}}{k q} o r t_{1} α r_{A} r_{B}$
Similarly $t_{2} α r_{B} r_{C}$
$∴ t_{1} < t_{2}$

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