Electric Potential on the surface of a charged spherical shell of radius 10 cm is 50 volt. Find the value of electric potential at a distance of 20 cm from the center of spherical shell.

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Solution

Potential at surface = 50 V
$V = \frac{K . Q}{R}$
$50 = \frac{K . Q}{0.1}$
KQ= 5
potential at 200 m from centre
$V = \frac{K . Q}{r}$
$V = \frac{5}{0.2}$ = 25 volt

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Electric Potential on the surface of a charged spherical shell of radius 10 cm is 50 volt. Find the value of electric potential at a distance of 20 cm from the center of spherical shell.

Potential at surface = 50 V V=K.QR50=K.Q0.1KQ= 5 potential at 200 m from centre V=K.QrV=50.2 = 25 volt

Potential at surface = 50 V
$V = \frac{K . Q}{R}$
$50 = \frac{K . Q}{0.1}$
KQ= 5
potential at 200 m from centre
$V = \frac{K . Q}{r}$
$V = \frac{5}{0.2}$ = 25 volt