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Question

Two charges 10μC and 5μC are at Im apart. In order to make this distance 0.5m, how much work would be done?

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Solution

q1=10μC=10×106Cq2=5μC=5×106C14πε0=k=9×109Nm2C2r1=1mr2=0.5m
First situation U1=kq1q2r1
Second situation U2=kq1q2r2
Work done W=U2U1
W=kq1q2r2kq1q2r1W=kq1q2[1r21r1]W=9×109×10×106×5×106[10.511]
W=45×102[10511]
W=45×102[21]
W=45×102J

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