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Question

(C) For a charged conducting sphere,the potential inside the sphere is constant and equal to the potential at its surface. Given radius $R = 10 \, cm$

Vedclass · Accepted Answer

$2/3 \, V$

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(C) For a charged conducting sphere,the potential inside the sphere is constant and equal to the potential at its surface. Given radius $R = 10 \, cm$. The potential at any point inside the sphere (including the center) is equal to the potential at the surface: $V_{surface} = \frac{KQ}{R} = \frac{KQ}{10} = V$. Thus,$KQ = 10V$. For a point outside the sphere at a distance $r = 15 \, cm$,the potential is given by $V_{outside} = \frac{KQ}{r}$. Substituting the values: $V_{outside} = \frac{10V}{15} = \frac{2}{3}V$.

If the electric potential at a distance of $5 \, cm$ from the center of a charged conducting sphere of radius $10 \, cm$ is $V$,then what will be the potential at a point at a distance of $15 \, cm$ from the center?

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