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ololo11 [35]
3 years ago
15

A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s

traight line to a new position at a distance R away from its starting position. The final location of q is at a distance r2 from Q. 1)What is the change in the potential energy of charge q during this process
Physics
1 answer:
topjm [15]3 years ago
7 0

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

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