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

Therefore,

`V_(10)=2.34* 10^(6)\ V`

`V_(20)=1.17* 10^(6)\ V`

`V_(14)=1.67* 10^(6)\ V`

Step-by-step 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=(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=(9* 10^(9)* 26* 10^(-6))/(0.1)`

`V=2.34* 10^(6)\ V`

At r = 20 cm = 0.2 m

`V=(9* 10^(9)* 26* 10^(-6))/(0.2)`

`V=1.17* 10^(6)\ V`

At r = 14 cm = 0.14 m

`V=(9* 10^(9)* 26* 10^(-6))/(0.14)`

`V=1.67* 10^(6)\ V`

Therefore,

`V_(10)=2.34* 10^(6)\ V`

`V_(20)=1.17* 10^(6)\ V`

`V_(14)=1.67* 10^(6)\ V`