The change in the electric potential energy at the final given distance is 38.69 J.
Given;
- magnitude of the first charge, q₁ = 2.255 x 10⁻⁴ C
- magnitude of the second charge, q₂ = 7.62 x 10⁻⁶ C
- initial distance between the two charges, r₁ = 0.35 m
- final distance between the two charges, r₂ = 1.55 m
To find:
Change in the potential energy
The initial potential energy between the charges is calculated as;
[tex]V_1 = \frac{kq_1q_2}{r_1} \\\\where;\\\\k \ is \ Coulomb's\ constant = 9\times 10^9 \ Nm/c^2\\\\V_1 = \frac{(9\times 10^9)\times (2.55\times 10^{-4}) \times (7.62 \times 10^{-6})}{0.35} \\\\V_1 = 49.97 \ J[/tex]
The final potential energy between the charges is calculated as;
[tex]V_2 = \frac{(9\times 10^9)\times (2.55\times 10^{-4}) \times (7.62 \times 10^{-6})}{1.55} \\\\V_2 = 11.28 \ J[/tex]
The change in the potential energy is calculated as;
ΔV = V₁ - V₂
= 49.97 J - 11.28 J
= 38.69 J
Therefore, the electric potential energy changed by 38.69 J
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