Question

Two point charges of \(1\,\text{nC}\) and \(2\,\text{nC}\) are placed at two corners of an equilateral triangle of side \(3\) cm. The work done in bringing a charge of \(3\,\text{nC}\) from infinity to the third corner of the triangle is ________ \(\mu\text{J}\). \[ \left(\frac{1}{4\pi\varepsilon_0}=9\times10^9\,\text{N m}^2\text{C}^{-2}\right) \]

• A. \(5.4\)
• B. \(27\)
• C. \(3.3\)
• D. \(2.7\)

Hint:

To find work done in electrostatics, always compute the electric potential first—this avoids dealing directly with forces.

Answer 1

The Correct Option is D

Concept: The work done in bringing a charge from infinity to a point in an electric field is equal to: \[ W = qV \] where \(q\) is the charge being brought and \(V\) is the electric potential at that point due to the existing charges. Electric potential due to a point charge: \[ V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} \]
Step 1: Identify given quantities \[ q_1 = 1\,\text{nC}, \quad q_2 = 2\,\text{nC}, \quad q = 3\,\text{nC} \] \[ r = 3\,\text{cm} = 0.03\,\text{m} \]
Step 2: Calculate the electric potential at the third corner Since the triangle is equilateral, the distance of the third corner from both charges is the same. \[ V = \frac{1}{4\pi\varepsilon_0} \left(\frac{q_1}{r} + \frac{q_2}{r}\right) = \frac{1}{4\pi\varepsilon_0}\frac{q_1+q_2}{r} \] \[ V = 9\times10^9 \times \frac{(1+2)\times10^{-9}}{0.03} \] \[ V = 9\times10^9 \times \frac{3\times10^{-9}}{0.03} = \frac{27}{0.03} = 900\ \text{V} \]
Step 3: Calculate the work done in bringing the charge \[ W = qV = 3\times10^{-9} \times 900 = 2700\times10^{-9}\ \text{J} = 2.7\times10^{-6}\ \text{J} \] \[ W = 2.7\ \mu\text{J} \] Final Answer: \[ \boxed{2.7\ \mu\text{J}} \]