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?

ΔV = -1321.73V

Explanation:

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

        (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:

                (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:

           (3)

You replace the values of a and b:

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

car a and b are moving in opposite directions

plz mark.me the brainliest i rlly need this kind of yours in exchange

the right answer is 0.3 meters per second