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=10×10−6Cq2=5μC=5×10−6C14πε0=k=9×109N⋅m2C2r1=1mr2=0.5m First situation U1=kq1q2r1 Second situation U2=kq1q2r2 Work done W=U2−U1 W=kq1q2r2−kq1q2r1W=kq1q2[1r2−1r1]W=9×109×10×10−6×5×10−6[10.5−11] W=45×10−2[105−11] W=45×10−2[2−1] W=45×10−2J