Question

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?

Answer 1

Explanation:

The given data is as follows.

     $r_{1}$ = 0.25 m,    q = 5.90 mC = $5.90 \times 10^{-3} C$

     $r_{2}$ = 0.35 m,    q = 1.70 mC = $1.70 \times 10^{-3} C$

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

             V = $k \frac{q}{r}$

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

    $V_{i} = 9 \times 10^{9} \times \frac{5.90 \times 10^{-3}}{0.25 m}$      

                = $212.4 \times 10^{6}$

or,             = $2.124 \times 10^{8}$

$V_{f} = 9 \times 10^{9} \times \frac{1.70 \times 10^{-3}}{0.35 m}$      

                = $43.71 \times 10^{6}$

or,             = $4.371 \times 10^{8}$

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

              $\Delta V = V_{f} - V_{i}$

                         = $4.371 \times 10^{8} - 2.124 \times 10^{8}$

                        = $2.247 \times 10^{8}$ V

Therefore, the difference in electric potential energy is $2.247 \times 10^{8}$ V.

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

                P.E = qV

                    = $-1.70 \times 10^{-3}C \times 2.247 \times 10^{8}$ V

                    = $-3.8199 \times 10^{5}$

Therefore, the change in the system's electric potential energy is $-3.8199 \times 10^{5}$.