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 required to move a charge from one point to another in the electric field created by a source charge at the origin.
Since the electric field is a conservative field, the work done is path-independent and equals the charge being moved multiplied by the change in electric potential.
Step 2: Key Formula or Approach:
1. The electric potential \(V\) at a distance \(r\) from a point source charge \(Q\) is given by the formula \(V = k\frac{Q}{r}\), where \(k = \frac{1}{4\pi\epsilon_0}\).
2. The work done \(W\) in moving a test charge \(q\) from an initial point A to a final point B is \(W = q \times (V_B - V_A)\), where \(V_A\) and \(V_B\) are the potentials at points A and B, respectively.
3. The distance \(r\) of a point \((x,y,z)\) from the origin is calculated using the distance formula: \(r = \sqrt{x^2+y^2+z^2}\).
Step 3: Detailed Explanation:
Given data:
Source charge, \(Q = 10^{-8}\) C.
Test charge, \(q = 2 \mu C = 2 \times 10^{-6}\) C.
Coulomb's constant, \(k = 9 \times 10^9\) N m\(^2\)/C\(^2\).
Initial point, A = (4, 4, 2) m.
Final point, B = (2, 2, 1) m.
First, we calculate the distances of points A and B from the origin.
Distance of A from origin: \(r_A = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16+16+4} = \sqrt{36} = 6\) m.
Distance of B from origin: \(r_B = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3\) m.
Next, we calculate the electric potential at each point.
Potential at A: \(V_A = k\frac{Q}{r_A} = (9 \times 10^9) \frac{10^{-8}}{6} = \frac{90}{6} = 15\) Volts.
Potential at B: \(V_B = k\frac{Q}{r_B} = (9 \times 10^9) \frac{10^{-8}}{3} = \frac{90}{3} = 30\) Volts.
Finally, we calculate the work done in moving the charge \(q\) from A to B.
\[ W_{A \to B} = q(V_B - V_A)
\] \[ W_{A \to B} = (2 \times 10^{-6} \text{ C}) \times (30 \text{ V} - 15 \text{ V})
\] \[ W_{A \to B} = (2 \times 10^{-6}) \times (15) = 30 \times 10^{-6} \text{ J}.
\] Step 4: Final Answer:
The work done in moving the charge from point A to point B is \(30 \times 10^{-6}\) J.