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

Step 1: Understanding the Concept:
Work done in bringing a charge from infinity is equal to the electric potential energy of the charge at that location.
Step 2: Key Formula or Approach:
Electric Potential \(V = \frac{1}{4\pi\epsilon_0} \sum \frac{q_i}{r_i}\).
Work Done \(W = q \cdot V\).
Step 3: Detailed Explanation:
1. Calculate Potential (\(V\)) at the third corner:
Potential due to \(1\ \text{nC}\) and \(2\ \text{nC}\) at distance \(3\ \text{cm} = 0.03\ \text{m}\):
\[ V = 9 \times 10^9 \cdot \left( \frac{1 \times 10^{-9}}{0.03} + \frac{2 \times 10^{-9}}{0.03} \right) \] \[ V = \frac{9 \times 10^9 \cdot 3 \times 10^{-9}}{0.03} = \frac{27}{0.03} = 900\ \text{V} \] 2. Calculate Work Done (\(W\)):
Using \(W = q_{test} \cdot V\) for \(q_{test} = 3\ \text{nC} = 3 \times 10^{-9}\ \text{C}\):
\[ W = 3 \times 10^{-9} \cdot 900 = 2700 \times 10^{-9}\ \text{J} = 2.7 \times 10^{-6}\ \text{J} \] \[ W = 2.7\ \mu\text{J} \] Step 4: Final Answer:
The work done is \(2.7\ \mu\text{J}\).