A 2.55x10^-4 C charge is held 0.350 m from a second 7.62x10^-6 C charge. The second one is released. When it reaches a distance of 1.55 m from the other charge, how much has the electric potential energy changed?
(Unit=J)

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 Cmagnitude of the second charge, q₂ = 7.62 x 10⁻⁶ Cinitial distance between the two charges, r₁ = 0.35 mfinal 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;

The final potential energy between the charges is calculated as;

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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