Two conducting spheres S1 and S2 of radii a and b (b > a) respectively, are placed far apart and connected by a long, thin conducting wire, as shown in the figure.

For some charge placed on this structure, the potential and surface electric field on S1 are $V_{a} a n d E_{a}$ , and that on S2 are $V_{b} a n d E_{b}$ respectively. Then, which of the following is CORRECT?

V_{a} = V_{b} a n d E_{a} < E_{b}

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V_{a} > V_{b} a n d E_{a} > E_{b}

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V_{a} = V_{b} a n d E_{a} > E_{b}

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V_{a} > V_{b} a n d E_{a} = E_{b}

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Solution

The correct option is C $V_{a} = V_{b} a n d E_{a} > E_{b}$
Ans : (c)

When charge is placed on this structure, equillibrium is established such the both spheres are at same potential i.e.

$V_{a} = V_{b}$

$V_{a} = V_{b}$

So, $\frac{Q_{a}}{4 π \in_{0} a} = \frac{Q_{b}}{4 π \in_{0} b}$
$\frac{Q_{b}}{Q_{a}} = \frac{b}{a}$

Now, surface electric fields,

$\frac{E_{a}}{E_{b}} = [\frac{Q_{a} / 4 π ε_{0} a^{2}}{Q_{b} / 4 π ε_{0} b^{2}}] = \frac{Q_{a} \times b^{2}}{Q_{b} \times a^{2}} = \frac{b}{a} > 1$

$S o, E_{a} > E_{b}$

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