Explanation:
The given data is as follows.
[tex]r_{1}[/tex] = 0.25 m, q = 5.90 mC = [tex]5.90 \times 10^{-3} C[/tex]
[tex]r_{2}[/tex] = 0.35 m, q = 1.70 mC = [tex]1.70 \times 10^{-3} C[/tex]
(a) Now, we will calculate the electric potential as follows.
V = [tex]k \frac{q}{r}[/tex]
First, we will calculate the initial and final electric potential as follows.
[tex]V_{i} = 9 \times 10^{9} \times \frac{5.90 \times 10^{-3}}{0.25 m}[/tex]
= [tex]212.4 \times 10^{6}[/tex]
or, = [tex]2.124 \times 10^{8}[/tex]
[tex]V_{f} = 9 \times 10^{9} \times \frac{1.70 \times 10^{-3}}{0.35 m}[/tex]
= [tex]43.71 \times 10^{6}[/tex]
or, = [tex]4.371 \times 10^{8}[/tex]
Hence, the value of change in electric potential is as follows.
[tex]\Delta V = V_{f} - V_{i}[/tex]
= [tex]4.371 \times 10^{8} - 2.124 \times 10^{8}[/tex]
= [tex]2.247 \times 10^{8}[/tex] V
Therefore, the difference in electric potential energy is [tex]2.247 \times 10^{8}[/tex] V.
(b) Now, we will calculate the potential energy as follows.
P.E = qV
= [tex]-1.70 \times 10^{-3}C \times 2.247 \times 10^{8}[/tex] V
= [tex]-3.8199 \times 10^{5}[/tex]
Therefore, the change in the system's electric potential energy is [tex]-3.8199 \times 10^{5}[/tex].
