Question

The work done to keep three charges \( 2 \times 10^{-5} \text{ C} \), \( 3 \times 10^{-5} \text{ C} \), \( 4 \times 10^{-5} \text{ C} \) at vertices of an equilateral triangle of side 10 cm is:

• A. \( 324 \text{ J} \)
• B. \( 234 \text{ J} \)
• C. \( 432 \text{ J} \)
• D. \( 224 \text{ J} \)

Hint:

When calculating work done in assembling charges, ensure the distance \( r \) is in meters.

Answer 1

The Correct Option is B

Concept: The work done to assemble a system of point charges is equivalent to the total electrostatic potential energy of the system. For three charges, this is the sum of the potential energies of all unique pairs: \( U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right) \), where \( k = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2} \).

Step 1: Identify all given values.
Charges: \( q_1 = 2 \times 10^{-5} \text{ C} \),
\( q_2 = 3 \times 10^{-5} \text{ C} \),
\( q_3 = 4 \times 10^{-5} \text{ C} \).
Distance \( r = 10 \text{ cm} = 0.1 \text{ m} \).
Since it is an equilateral triangle, all distances
\( r_{12} = r_{23} = r_{31} = r = 0.1 \text{ m} \).

Step 2: Calculate potential energy for each pair.
The work done is: $$ W = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r} + \frac{q_2 q_3}{r} + \frac{q_3 q_1}{r} \right) $$ $$ W = \frac{9 \times 10^9}{0.1} \left( (2 \times 10^{-5} \times 3 \times 10^{-5}) + (3 \times 10^{-5} \times 4 \times 10^{-5}) + (4 \times 10^{-5} \times 2 \times 10^{-5}) \right) $$

Step 3: Evaluate the numerical expression.
$$ W = 9 \times 10^{10} \times 10^{-10} \left( (2 \times 3) + (3 \times 4) + (4 \times 2) \right) $$ $$ W = 9 \times (6 + 12 + 8) $$ $$ W = 9 \times 26 = 234 \text{ J} $$ $$\boxed{234 \text{ J}}$$