Question

Two positive point charges are of $12\mu C$ and $8\mu C$ are $10cm$ apart from each other. The work done in bringing them $4cm$ closer is

• A. 5.8 J
• B. 13 eV
• C. 5.8 eV
• D. 13 J

Answer 1

The correct option is D

13 J

Work done:

The electric potential at any point is defined as the amount of work done in bringing a unit positive charge from infinity to that point.

1. When two attracting charges are brought closer together, the potential energy decreases.
2. Work done is expressed as the difference between the potentials in moving a charge from one point to another.

Work done in bringing charges together is equal to the difference in potential which is a function of charges (q) and distance (d) between them.

Step 1: Given data

Distance, $$\begin{gathered} d_1=4cm \\ {d}_2=10cm \end{gathered}$$

Charges, $$\begin{gathered} q_1=12\mu C \\ {q}_2=8\mu C \end{gathered}$$

Step 2: Formula used

$$\begin{gathered} Workdone=V_1-V_2 \\ Potentialbetweenchargesaddis\tan ced=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d} \\ Workdone=\frac{1}{4\pi {\epsilon }_0}{q}_1{q}_2\left(\frac{1}{d_1}-\frac{1}{d_2}\right) \end{gathered}$$

Step 3: Calculation

Substituting the values, we get

$$\begin{gathered} W=9\times {10}^9\times 12\times {10}^{-6}\times 8\times {10}^{-6}\left(\frac{1}{0.04}-\frac{1}{0.1}\right) \\ \left(\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\right) \\ W=129.6\times 10 \\ =12.96J \end{gathered}$$

That shows the value of work done in nearly $13joules$.

Hence, option D is correct.