Answer to Question #305960 in Solid State Physics for Yhel

Question

How much work is needed to decrease the distance between a +15 µC charge and a -20 µC charge from 1 m to 0.25 m?

Answer 1

The potential energy of the first configuration:

$$U_1 = k\dfrac{q_1q_2}{r_1}$$

where $q_1 = +15\times 10^{-6}C, q_2 = -20\times 10^{-6}C,r_1 = 1m$.

The potential energy of the second configuration:

$$U_2 = k\dfrac{q_1q_2}{r_2}$$

where $r_2 = 0.25m$.

The work is the difference between these potential energies:

$$W =| U_2-U_1| = |kq_1q_2(1/r_2 - 1/r_1)| = \\ =9\times10^9\cdot 15\cdot 20\cdot (10^{-6})^2\cdot (1/0.25 - 1/1)=8.1J$$

Answer. 8.1J.