First slide

Electrostatic potential

Question

Two spheres of radius a and b respectively are charged and joined by a wire. The ratio of electric field of the spheres is

Easy

A

$a / b$
B

$b / a$
C

$a^{2} / b^{2}$
D

$b^{2} / a^{2}$

Solution

Joined by a wire means they are at the same potential. For same potential $\frac{{kQ}_{1}}{a_{1}} = \frac{{kQ}_{2}}{a_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{a}{b}$
Further, the electric field at the surface of the sphere having radius R and charge Q is $\frac{kQ}{R^{2}} .$
$∴ \frac{E_{1}}{E_{2}} = \frac{{kQ}_{1} / a^{2}}{{kQ}_{2} / b_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{b^{2}}{a^{2}} = \frac{b}{a}$

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