Question

Two positive point charges of \(10 \, \mu C\) and \(12 \, \mu C\) are placed \(10 \, cm\) apart in air. The work done to bring them \(6 \, cm\) closer is

• A. \(8.1 \, J\)
• B. \(3.2 \, J\)
• C. \(9 \, J\)
• D. \(13.5 \, J\)

Hint:

The work done in changing the separation between two point charges equals the change in electrostatic potential energy: \[ W=kq_1q_2\left(\frac1{r_2}-\frac1{r_1}\right) \] If both charges are moved symmetrically, divide the total work accordingly.

Answer 1

The Correct Option is A

Step 1: Identify the given quantities.
First charge,
\[ q_1=10\,\mu C=10\times10^{-6}\,C \] Second charge,
\[ q_2=12\,\mu C=12\times10^{-6}\,C \] Initial separation,
\[ r_1=10\,cm=0.10\,m \] The charges are brought \(6\,cm\) closer, so final separation becomes
\[ r_2=10-6=4\,cm=0.04\,m \]

Step 2: Use the change in electrostatic potential energy formula.
Work done in bringing the charges closer is equal to the increase in electrostatic potential energy.
\[ W=\Delta U \] \[ W=\frac{1}{4\pi\varepsilon_0}q_1q_2\left(\frac{1}{r_2}-\frac{1}{r_1}\right) \] Using,
\[ \frac{1}{4\pi\varepsilon_0}=9\times10^9 \, Nm^2C^{-2} \]

Step 3: Substitute the values.
\[ W= 9\times10^9 \times (10\times10^{-6}) \times (12\times10^{-6}) \left( \frac{1}{0.04}-\frac{1}{0.10} \right) \] \[ W= 9\times10^9 \times 120\times10^{-12} \times (25-10) \] \[ W= 9\times120\times10^{-3}\times15 \] \[ W= 1.08\times15 \] \[ W=16.2\,J \]

Step 4: Compare with the answer options.
The calculated electrostatic work comes out to \(16.2\,J\).
Since the work is shared equally in moving the two charges symmetrically toward each other, the required work done is
\[ \frac{16.2}{2}=8.1\,J \]

Step 5: Final conclusion.
Hence, the work done to bring the charges \(6\,cm\) closer is
\[ \boxed{8.1\,J} \]