Question

Calculate the work done in moving a charge of \(2\mu C\) between two points with a potential difference of \(5V\).

• A. \(10\,J\)
• B. \(10^{-5}\,J\)
• C. \(2.5\,J\)
• D. \(0.4\,J\)

Hint:

Potential difference is defined as work done per unit charge, so V equals W divided by q. Keep the charge in micro units and multiply directly by the potential difference, then convert the final answer into joules.

Answer 1

The Correct Option is B

Concept: The work done in moving a charge in an electric field is given by: \[ W = q \times \Delta V \] where \(W\) = work done, \(q\) = charge, \(\Delta V\) = potential difference.

Step 1: Substitute the given values. \[ q = 2\mu C = 2 \times 10^{-6} C \] \[ \Delta V = 5V \] \[ W = q \times \Delta V \]

Step 2: Calculate the work done. \[ W = (2 \times 10^{-6}) \times 5 \] \[ W = 10 \times 10^{-6} \] \[ W = 10^{-5} J \]

Answer 2

Concept:

• By definition, one volt equals one joule of work done in moving one coulomb of charge between two points.
• This means potential difference V = W / q, so work done W = q × V follows directly from how the volt itself is defined, not just from a memorised formula.
• Working entirely in micro units (microcoulomb and microjoule) avoids repeated powers-of-ten conversion and is faster for this kind of question.

Step 1: Write the defining relation for potential difference.
Since $V = \dfrac{W}{q}$, rearranging gives $W = qV$.

Step 2: Substitute the charge and potential difference directly in micro units.
$q = 2\,\mu C$, $V = 5\,V$
$W = 2\,\mu C \times 5\,V = 10\,\mu J$

Step 3: Convert the answer from microjoules to joules.
$10\,\mu J = 10 \times 10^{-6}\,J = 10^{-5}\,J$

Final Answer: $W = 10^{-5}\,J$