Question

A nonconducting sphere of radius 10 cm is charged uniformly with a density of 100 nC/m3. What is the magnitude of the potential difference between the center and a point 4.0 cm away?

Answer 1

Step-by-step explanation:

The given data is as follows.

Radius of the sphere (R) = 10 cm = 0.01 m (as 1 m = 100 cm)

Distance from the center (r) = 4 cm = 0.04 m

Charge density (
`\sigma`) = 100
`nC/m^(3)`

=
`100 * 10^(-9) C/m^(3)` (As 1 nm =
`10^(-9) m`)

As the relation between charge and potential difference is as follows.

Q =
`\sigma V`

=
`\sigma ((4)/(3) \pi r^(3))`

=
`100 * 10^(-9) C/m^(3) * (4)/(3) \pi (0.01)^(3)`

=
`4.19 * 10^(-10) C`

Expression for electric field is as follows.

E(r) =
`k (qr)/(R^(3))`

Electric potential, V(r) =
`-\int_(0)^(r) E(r) dr`

=
`-\int_(0)^(r) k (qr)/(R^(3)) dr`

=
`-k ((qr^(2))/(2R^(3)))`

=
`-9 * 10^(9) Nm^(2)/C^(2) ((4.19 * 10^(-10) C (0.04)^(2))/(2(0.1)^(3)))`

= -3 V

Thus, we can conclude that the magnitude of the potential difference between the center and a point 4.0 cm away is -3 V.