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Q.

Two spherical conductors of radii $R_{1}$ and $R_{2}$ are separated by a distance much larger than the radius of either sphere. The spheres are connected by a conducting wire as shown in Fig. If the charges on the spheres in equilibrium are $q_{1}$ and $q_{2}$ , respectively, what is the ratio of the field strength at the surfaces of the spheres?

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a

$R_{2} / R_{1}$

b

$R_{2}^{2} / R_{1}^{2}$

c

$R_{1} / R_{2}$

d

$R_{1}^{2} / R_{2}^{2}$

answer is A.

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Detailed Solution

Their potential will be same, i.e., $V_{1} = V_{2}$

$\frac{k q_{1}}{R_{1}} \frac{k q_{2}}{R_{2}} o r \frac{q_{1}}{q_{2}} = \frac{R_{1}}{R_{2}}$ $\Rightarrow \frac{E_{1}}{E_{2}} = \frac{k q_{1} / R_{1}^{2}}{k q_{2} / R_{2}^{2}} = \frac{q_{1}}{q_{2}} {(\frac{R_{2}}{R_{1}})}^{2}$ $= \frac{R_{1}}{R_{2}} {(\frac{R_{2}}{R_{1}})}^{2} = \frac{R_{2}}{R_{1}}$

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