Answer to Question #305960 in Solid State Physics for Yhel

Question #305960

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?


1
Expert's answer
2022-03-04T13:22:58-0500

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)| = \\\\\n=9\\times10^9\\cdot 15\\cdot 20\\cdot (10^{-6})^2\\cdot (1\/0.25 - 1\/1)=8.1J"

Answer. 8.1J.


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