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(C) The electric potential $V$ of a charged spherical conductor (like a soap bubble) is given by the formula $V = \frac{1}{4\pi\epsilon_0} \cdot \frac

Vedclass · Accepted Answer

$8$

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(C) The electric potential $V$ of a charged spherical conductor (like a soap bubble) is given by the formula $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r}$,where $q$ is the charge and $r$ is the radius.Since the charge $q$ on the soap bubble remains constant when its radius is changed,we have $V \propto \frac{1}{r}$.Given the initial potential $V_1 = 16\,V$ and initial radius $r_1 = r$.The new radius is $r_2 = 2r$.Using the proportionality $V_1 r_1 = V_2 r_2$,we get:$16 	imes r = V_2 	imes (2r)$$V_2 = \frac{16}{2} = 8\,V$.Therefore,the new potential of the bubble will be $8\,V$.

The radius of a soap bubble whose potential is $16\,V$ is doubled. The new potential of the bubble will be.....$V$

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