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:


U1=kq1q2r1U_1 = k\dfrac{q_1q_2}{r_1}

where q1=+15×106C,q2=20×106C,r1=1mq_1 = +15\times 10^{-6}C, q_2 = -20\times 10^{-6}C,r_1 = 1m.

The potential energy of the second configuration:


U2=kq1q2r2U_2 = k\dfrac{q_1q_2}{r_2}

where r2=0.25mr_2 = 0.25m.

The work is the difference between these potential energies:


W=U2U1=kq1q2(1/r21/r1)==9×1091520(106)2(1/0.251/1)=8.1JW =| 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.


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