Question

Charge Q = +4.00 mC is distributed uniformly over the volume of an insulating sphere that has radius R = 5.00 cm. What is the potential difference between the center of the sphere and the surface of the sphere?

Answer 1

Answer:

3.6 x 10^8 V

Explanation:

Q = 4 m C = 4 x 10^-3 C

r = 5 cm = 0.05 m

The formula for the potential at the surface is

Vs = K Q / r = (9 x 10^9 x 4 x 10^-3) / 0.05 = 7.2 x 10^8 V

The formula for the potential at the centre is

Vc = 3/2 Vs

Vc = 1.5 x 7.2 x 10^8 V = 10.8 x 10^8 V

The difference in potential is

V = Vc - Vs = 10.8 x 10^8 - 7.2 x 10^8 = 3.6 x 10^8 V

Answer 2

The potential difference between the center of the sphere and the surface of the sphere will be PD=3.6 x 10^8 V

What will be the potential difference?

It is given that:-

Charge Q = +4.00 mC  or  $4\times10^{-3} c$

insulating sphere radius R = 5.00 cm = 0.05 m

The formula for the potential difference at the surface of the sphere is given by

$V_{s} =\dfrac{K\times Q}{r} =\dfrac{9\times10^{9} \times 4\times10^{-3} }{0.05}=7.2\times10^{8} V$  

The formula for the potential difference at the center of the sphere  

$Vc_{} =\dfrac{3}{2} \times V_{s}$

$V_{c} = 1.5\times7.2\times10^{8} v = 10.8\times10^{8 } V$      

The difference between the potential difference will be  

$V=V_{c} -V_{s} = 10.8\times10^{8} =3.6\times10^{8} V$

Thus the potential difference between the center of the sphere and the surface of the sphere will be PD=3.6 x 10^8 V

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