Question

Two point charges $+10\mu C$ and $4\mu C$ are placed 10 cm apart in air. The work required to be done to bring them 2 cm closer is ($\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ SI units)

• A. 0.65 J
• B. 0.9 J
• C. 1.2 J
• D. 2.3 J

Hint:

Work is positive when bringing like charges closer, as you are moving against the electrostatic repulsion.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Work done in moving a charge in an electric field is equal to the change in electrostatic potential energy of the system. Bringing them "2 cm closer" means the final distance is $10 - 2 = 8$ cm.
Step 2: Key Formula or Approach:
Electrostatic potential energy $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} = k \frac{q_1 q_2}{r}$
Work done $W = \Delta U = U_{final} - U_{initial} = k q_1 q_2 \left(\frac{1}{r_f} - \frac{1}{r_i}\right)$
Step 3: Detailed Explanation:
Given:
$q_1 = 10 \times 10^{-6} C$
$q_2 = 4 \times 10^{-6} C$
$r_i = 10 cm = 0.1 m$
$r_f = 10 - 2 = 8 cm = 0.08 m$
$k = 9 \times 10^9$
Substitute the values:
\[ W = 9 \times 10^9 \times (10 \times 10^{-6}) \times (4 \times 10^{-6}) \left(\frac{1}{0.08} - \frac{1}{0.1}\right) \]
\[ W = 9 \times 10^9 \times 40 \times 10^{-12} \left(12.5 - 10\right) \]
\[ W = 360 \times 10^{-3} \times (2.5) \]
\[ W = 0.36 \times 2.5 = 0.9 J \]
Step 4: Final Answer:
The work required is 0.9 J.