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If a conducting sphere of radius $R$ is charged. Then the electric field at a distance $r(r>R)$ from the centre of the sphere would be, $(V=$ potential on the surface of the sphere $)$
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Verified Answer
The correct answer is:
$\frac{R V}{r^2}$
For $r>R$,
Electric field $E=\frac{K q}{r^2} \ldots$
and potential at surface of sphere $V=\frac{K q}{R}$
$\therefore$ From equation (1)
$V=\frac{E r^2}{R}$
$\Rightarrow E=\frac{R V}{r^2}$
Electric field $E=\frac{K q}{r^2} \ldots$
and potential at surface of sphere $V=\frac{K q}{R}$
$\therefore$ From equation (1)
$V=\frac{E r^2}{R}$
$\Rightarrow E=\frac{R V}{r^2}$
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