Two charged conducting spheres of radii $r_{1}$ and $r_{2}$ have same electric field near their surfaces. The ratio of their electric potential is

r_{1}^{2} / r_{2}^{2}

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r_{2}^{2} / r_{1}^{2}

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r_{1} / r_{2}

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r_{2} / r_{1}

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Solution

The correct option is C $r_{1} / r_{2}$

Given,

$E_{A} = E_{B}$

Using the formula of electric field near the surface of sphere,

$\Rightarrow \frac{k q_{1}}{r_{1}^{2}} = \frac{k q_{2}}{r_{2}^{2}}$

$\Rightarrow \frac{q_{1}}{q_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}} . . . . . . . . (1)$

So potential of both spheres will be

$V_{A} = \frac{k q_{1}}{r_{1}}; V_{B} = \frac{k q_{2}}{r_{2}}$

$\Rightarrow \frac{V_{A}}{V_{B}} = \frac{q_{1}}{r_{1}} \cdot \frac{r_{2}}{q_{2}} = \frac{q_{1}}{q_{2}} \cdot \frac{r_{2}}{r_{1}}$

From eq. $(1)$

$\Rightarrow \frac{V_{A}}{V_{B}} = \frac{r_{1}^{2}}{r_{2}^{2}} \cdot \frac{r_{2}}{r_{1}} = \frac{r_{1}}{r_{2}}$

Hence, option $(c)$ is correct.

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