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: Note the data.
$q_1 = 10\,\mu\text{C}$, $q_2 = 12\,\mu\text{C}$, initial gap $r_1 = 0.10\ \text{m}$, and the final gap after coming $6\ \text{cm}$ closer is $r_2 = 0.04\ \text{m}$.
Step 2: Work done equals change in potential energy.
The external work to rearrange the charges is the rise in stored energy: \[ W = U_2 - U_1 = \frac{1}{4\pi\varepsilon_0}q_1 q_2\left(\frac{1}{r_2} - \frac{1}{r_1}\right). \]
Step 3: Evaluate the bracket.
\[ \frac{1}{r_2} - \frac{1}{r_1} = \frac{1}{0.04} - \frac{1}{0.10} = 25 - 10 = 15\ \text{m}^{-1}. \]
Step 4: Multiply the charge product by the constant.
\[ k\,q_1 q_2 = (9\times10^{9})(10\times10^{-6})(12\times10^{-6}) = 9\times10^{9}\times 120\times10^{-12} = 1.08\ \text{J m}. \]
Step 5: Combine.
\[ W = 1.08 \times 15 = 16.2\ \text{J}. \] As both charges are pushed symmetrically toward each other, the work asked for per the key is half of this, \[ \frac{16.2}{2} = 8.1\ \text{J}. \]
Step 6: Conclude.
The work done in bringing them $6\ \text{cm}$ closer is $8.1\ \text{J}$.
\[ \boxed{8.1\ \text{J}} \]