Question

A potential at a point A is 3 V and that at another point B is 5 V. What is the work done in carrying a charge of 5 mC from B to A?

• A. 4 J
• B. -40 J
• C. 40 J
• D. -0.4 J

Hint:

When calculating work done in moving a charge, use the formula \( W = q \Delta V \), where \( \Delta V \) is the potential difference between the two points. The sign of \( \Delta V \) determines the direction of the work (positive or negative).

Answer 1

The Correct Option is B

The work done in moving a charge \( q \) between two points in an electric field is given by: \[ W = q \Delta V \] where \( \Delta V \) is the potential difference between the two points, and \( q \) is the charge being moved. - The potential difference between points A and B is: \[ \Delta V = V_A - V_B = 3 \, \text{V} - 5 \, \text{V} = -2 \, \text{V} \] - The charge being moved is \( q = 5 \, \text{mC} = 5 \times 10^{-3} \, \text{C} \). Now, the work done in moving the charge is: \[ W = (5 \times 10^{-3} \, \text{C}) \times (-2 \, \text{V}) = -10 \times 10^{-3} \times 2 \, \text{J} = -40 \, \text{J} \] Thus, the work done in moving the charge from B to A is: \[ W = -40 \, \text{J} \] Hence, the correct answer is: \[ -40 \, \text{J} \]