Question:
The work done in assembling three charged particles A, B and C of charges \(2 \times 10^{-5} C\), \(3 \times 10^{-5} C\) and \(4 \times 10^{-5} C\) respectively at the vertices of an equilateral triangle of side 10 cm is \([ \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 ]\).
The work done in assembling three charged particles A, B and C of charges \(2 \times 10^{-5} C\), \(3 \times 10^{-5} C\) and \(4 \times 10^{-5} C\) respectively at the vertices of an equilateral triangle of side 10 cm is \([ \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 ]\).
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Electrostatic assembly energy = sum of all pairwise interaction energies.
Updated On: Jun 20, 2026
- 126 J
- 180 J
- 234 J
- 162 J
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The Correct Option is C
Solution and Explanation
Step 1: Concept of electrostatic potential energy.
Work done in assembling charges equals total electrostatic potential energy: \[ W = \frac{1}{4\pi\varepsilon_0} \sum \frac{q_i q_j}{r} \] For three charges in equilateral triangle: \[ W = k \left(\frac{q_1q_2}{r} + \frac{q_2q_3}{r} + \frac{q_3q_1}{r}\right) \]
Step 2: Given values.
\[ q_1 = 2 \times 10^{-5}, \quad q_2 = 3 \times 10^{-5}, \quad q_3 = 4 \times 10^{-5} \] \[ r = 10 \text{ cm} = 0.1 \text{ m}, \quad k = 9 \times 10^9 \]
Step 3: Compute charge products.
\[ q_1q_2 = 6 \times 10^{-10} \] \[ q_2q_3 = 12 \times 10^{-10} \] \[ q_3q_1 = 8 \times 10^{-10} \]
Step 4: Sum of products.
\[ \sum q_i q_j = (6 + 12 + 8)\times 10^{-10} = 26 \times 10^{-10} = 2.6 \times 10^{-9} \]
Step 5: Substitute into formula.
\[ W = 9 \times 10^9 \times \frac{2.6 \times 10^{-9}}{0.1} \] \[ W = 9 \times 10^9 \times 2.6 \times 10^{-8} \] \[ W = 234 \, \text{J} \]
Step 6: Final interpretation.
Work done is positive because work is required to bring like charges from infinity to form the configuration: \[ \boxed{234 \, \text{J}} \]
Work done in assembling charges equals total electrostatic potential energy: \[ W = \frac{1}{4\pi\varepsilon_0} \sum \frac{q_i q_j}{r} \] For three charges in equilateral triangle: \[ W = k \left(\frac{q_1q_2}{r} + \frac{q_2q_3}{r} + \frac{q_3q_1}{r}\right) \]
Step 2: Given values.
\[ q_1 = 2 \times 10^{-5}, \quad q_2 = 3 \times 10^{-5}, \quad q_3 = 4 \times 10^{-5} \] \[ r = 10 \text{ cm} = 0.1 \text{ m}, \quad k = 9 \times 10^9 \]
Step 3: Compute charge products.
\[ q_1q_2 = 6 \times 10^{-10} \] \[ q_2q_3 = 12 \times 10^{-10} \] \[ q_3q_1 = 8 \times 10^{-10} \]
Step 4: Sum of products.
\[ \sum q_i q_j = (6 + 12 + 8)\times 10^{-10} = 26 \times 10^{-10} = 2.6 \times 10^{-9} \]
Step 5: Substitute into formula.
\[ W = 9 \times 10^9 \times \frac{2.6 \times 10^{-9}}{0.1} \] \[ W = 9 \times 10^9 \times 2.6 \times 10^{-8} \] \[ W = 234 \, \text{J} \]
Step 6: Final interpretation.
Work done is positive because work is required to bring like charges from infinity to form the configuration: \[ \boxed{234 \, \text{J}} \]
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