Question

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

Answer 1

Potential at any point inside the conductor will be same as surface
So V =K /r
Where k is constant and r is the distance from centre
So V =k/ 10
K =10V
Vat r=15=k/15=10V/15=2/3V

Answer 2

Answer: The potential at a point distance 15 cm from the center will be $\dfrac{2}{3}V$.

Explanation:

Given that,

Radius r = 10 cm

Potential at a point distance = 15 cm

If a charged spherical conductor of radius 10 cm has potential V at a point distance 5 cm from its center.

So, The potential V is

$V = \dfrac{kQ}{R}$

Where, R = radius of sphere

$V = \dfrac{kQ}{10}$......(I)

The potential at a point distance 15 cm from the center

The potential V'  is

$V' = \dfrac{kQ}{r}$

Where, r = point distance from the center

$V' = \dfrac{kQ}{15}$.....(II)

Divided equation (I) by equation(II)

$\dfrac{V}{V'}=\dfrac{\dfrac{kQ}{10}}{\dfrac{kQ}{15}}$

$\dfrac{V}{V'}=\dfrac{15}{10}$

$\dfrac{V}{V'}=\dfrac{3}{2}$

$V'=\dfrac{2}{3}V$

Hence, The potential at a point distance 15 cm from the center will be $\dfrac{2}{3}V$.