Question

Two positive charges of $12\mu C$ and $8\mu C$, respectively, are separated by $10cm$ apart in air. The work to be done to decrease the distance by $4cm$ is:

• A. Zero
• B. $388.8J$
• C. $459.8J$
• D. $5.76J$

Answer 1

The correct option is D $5.76J$

The potential energy of the system when the charges $Q_1$ and $Q_2$ are separated by a distance $d$ is given by

$U=\frac{1}{4\pi {ϵ}_0}\frac{Q_1{Q}_2}{d}$

Given $Q_1=12\mu C$ and $Q_2=8\mu C$

$d_i=10\text{cm}$ and $d_f=6\text{cm}$

Initial potential energy is $U_i=\frac{1}{4\pi {ϵ}_0}\frac{Q_1{Q}_2}{d_i}$ and

Final potential energy is $U_f=\frac{1}{4\pi {ϵ}_0}\frac{Q_1{Q}_2}{d_f}$

Thus, work done = change in energy i.e. $W=U_f-U_i=\frac{Q_1{Q}_2}{4\pi {ϵ}_0}(\frac{1}{d_f}-\frac{1}{d_i})$

Using $\frac{1}{4\pi {ϵ}_0}=9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$

$W=9\times {10}^9(12\times 8)\times {10}^{-12}(\frac{1}{0.06}-\frac{1}{0.1})=5.76J$