A conducting sphere with a radius of 0.25 m has a total charge of 5.90 mC. A particle with a charge of −1.70 mC is initially 0.35 m from the sphere's center and is moved to a final position 0.48 m from the sphere's center.

a. What is the difference in electric potential between the particle's final and initial positions, ΔV = Vf − Vi?
b. What is the change in the system's electric potential energy?

Explanation:

The given data is as follows.

      = 0.25 m,    q = 5.90 mC =

      = 0.35 m,    q = 1.70 mC =

(a)  Now, we will calculate the electric potential as follows.

             V =

First, we will calculate the initial and final electric potential as follows.

                =

or,             =

                =

or,             =

Hence, the value of change in electric potential is as follows.

                         =

                        = V

Therefore, the difference in electric potential energy is V.

(b)  Now, we will calculate the potential energy as follows.

                P.E = qV

                    = V

                    =

Therefore, the change in the system's electric potential energy is .

answer: decibel

explanation: