If a charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre, then the potential at a point distant 15 cm from the centre will be:

Question

If a charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre, then the potential at a point distant 15 cm from the centre will be: (A)  text3V  (B)  text3/2V  (C)  text2/3V  (D)  text1/3V

Tardigrade Admin · Accepted Answer

We know that the potential inside and on the surface of a spherical conductor is same so V=Vsurface=(q/10) textstat textvolt Vout=(q/15) textstat textvolt (Vout/V)=(2/3) so Vout=(2/3)V

Q. If a charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre, then the potential at a point distant 15 cm from the centre will be:

$3V$

$3/2V$

$2/3V$

$1/3V$

We know that the potential inside and on the surface of a spherical conductor is same so $V = V_{s u r f a ce} = \frac{q}{10} stat volt$ $V_{o u t} = \frac{q}{15} stat volt$ $\frac{V _{o u t}}{V} = \frac{2}{3}$ so $V_{o u t} = \frac{2}{3} V$

Q. If a charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre, then the potential at a point distant 15 cm from the centre will be:

3V

3/2V

2/3V

1/3V

We know that the potential inside and on the surface of a spherical conductor is same so V=Vsurface​=10q​statvolt Vout​=15q​statvolt VVout​​=32​ so Vout​=32​V

$3V$

$3/2V$

$2/3V$

$1/3V$

We know that the potential inside and on the surface of a spherical conductor is same so $V = V_{s u r f a ce} = \frac{q}{10} stat volt$ $V_{o u t} = \frac{q}{15} stat volt$ $\frac{V _{o u t}}{V} = \frac{2}{3}$ so $V_{o u t} = \frac{2}{3} V$