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

ΔV = -1321.73V

Step-by-step 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(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 (dr)/(r^2)=-kQ[-(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)[(1)/(0.055m)-(1)/(0.021m)]\\\\\Delta V=-1321.73V`

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