Question

35. A spherical conductor has a radius of 14.0 cm and a charge of 26.0 mC. Calculate the electric field and the electric poten- tial at (a) r 5 10.0 cm, (b) r 5 20.0 cm, and (c) r 5 14.0 cm from the center.

Answer 1

Answer:

Therefore,

$V_{10}=2.34\times 10^{6}\ V$

$V_{20}=1.17\times 10^{6}\ V$

$V_{14}=1.67\times 10^{6}\ V$

Explanation:

Given:

A spherical conductor has a radius of 14.0 cm

Q = 26.0 μC ( Assume it to be microC in the question it is miliC )

To Find:

Electric potential at

(a) r 5 10.0 cm, (b) r 5 20.0 cm, and (c) r 5 14.0 cm from the center.

Solution:

Electric potential due to point charge Q at any distance r from the charge is,

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

Where,

V = Electric Potential

k = Coulombs constant = 9 × 10⁹ Nm²/C².

Q = Charge

r = distance in meter

Substituting the values we get

At r = 10 cm = 0.1 m

$V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.1}$

$V=2.34\times 10^{6}\ V$

At r = 20 cm = 0.2 m

$V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.2}$

$V=1.17\times 10^{6}\ V$

At r = 14 cm = 0.14 m

$V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.14}$

$V=1.67\times 10^{6}\ V$

Therefore,

$V_{10}=2.34\times 10^{6}\ V$

$V_{20}=1.17\times 10^{6}\ V$

$V_{14}=1.67\times 10^{6}\ V$