Question

Location C is 0.021 m from a small sphere that has a charge of 5 nC uniformly distributed on its surface. Location D is 0.055 m from the sphere. What is the change in potential along a path from C to D?

Answer 1

Answer:

ΔV = -1321.73V

Explanation:

The change in potential along the path from C to D is given by the following expression:

$\Delta V=-\int_a^bE dr$         (1)

E: electric field produced by a charge at a distance of r

a: distance to the sphere at position C = 0.021m

b: distance to the sphere at position D = 0.055m

The electric field is given by:

$E=k\frac{Q}{r^2}$                 (2)

Q: charge of the sphere = 5nC = 5*10^-9C

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

You replace the expression (2) into the equation (1) and solve the integral:

$\Delta V=-kQ\int_a^b \frac{dr}{r^2}=-kQ[-\frac{1}{r}]_a^b$            (3)

You replace the values of a and b:

$\Delta V=(8.98*10^9Nm^2/C^2)(5*10^{-9}C)[\frac{1}{0.055m}-\frac{1}{0.021m}]\\\\\Delta V=-1321.73V$

The change in the potential along the path C-D is -1321.73V