Question

A point charge of \(10^{-8}\) C is placed at origin. The work done in moving a point charge 2 \(\mu\)C from point A(4, 4, 2) m to B(2, 2, 1) m is_______ J. (\(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\) in SI units)

• A. \(45 \times 10^{-6}\)
• B. 0
• C. \(30 \times 10^{-6}\)
• D. \(15 \times 10^{-6}\)

Hint:

The work done in a conservative field like the electric field only depends on the initial and final positions, not the path taken.
This allows us to use the potential difference, which is a much simpler calculation than integrating the force along a path.
Be careful with signs: work done by the field is \(q(V_A - V_B)\), while work done by an external agent to move the charge is \(q(V_B - V_A)\). The question implies the latter.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are asked to calculate the work done in moving a charge from point A to point B in the electric field of a point charge placed at the origin.
Since the electric field due to a point charge is conservative, the work done depends only on the initial and final positions and is equal to the charge multiplied by the change in electric potential.

Step 2: Relevant Formulae:
Electric potential due to a point charge: \[ V = k\frac{Q}{r} \] Work done in moving a charge: \[ W = q\,(V_B - V_A) \] Distance from origin: \[ r = \sqrt{x^2 + y^2 + z^2} \]
Step 3: Calculation:
Given:
Source charge, \(Q = 10^{-8}\,\text{C}\)
Test charge, \(q = 2\,\mu\text{C} = 2 \times 10^{-6}\,\text{C}\)
Coulomb constant, \(k = 9 \times 10^9\,\text{N m}^2\text{/C}^2\)
Point A = (4, 4, 2) m
Point B = (2, 2, 1) m

Distance of A from origin: \[ r_A = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{36} = 6\,\text{m} \] Distance of B from origin: \[ r_B = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3\,\text{m} \] Electric potential at A: \[ V_A = \frac{9 \times 10^9 \times 10^{-8}}{6} = 15\,\text{V} \] Electric potential at B: \[ V_B = \frac{9 \times 10^9 \times 10^{-8}}{3} = 30\,\text{V} \] Work done in moving the charge from A to B: \[ W = q (V_B - V_A) = (2 \times 10^{-6})(30 - 15) \] \[ W = 30 \times 10^{-6}\,\text{J} \] Step 4: Final Answer:
The work done in moving the charge from point A to point B is \[ \boxed{30 \times 10^{-6}\,\text{J}} \]