Question

A proton is taken from point P1 to point P2, both located in an electric field. The potentials at points P1 and P2 are -5 V and +5 V respectively. Assuming that kinetic energies of the proton at points P1 and P2 are zero, the work done on the proton is:

• A. \(1.6 \times 10^{-18} \, J}\)
• B. \(1.6 \times 10^{-18} \, J}\)
• C. Zero
• D. \(0.8 \times 10^{-18} \, J}\)

Hint:

Remember, the work done in moving a charge in an electric field is dependent on the potential difference and the charge itself. The direction of the electric field and the initial and final potentials will determine whether the work is done by the field or against it.

Answer 1

The Correct Option is B

Step 1: Calculating the potential difference. The potential difference \( V \) between points P1 and P2 is calculated as: \[ V = V_{P2}} - V_{P1}} = 5 \, V} - (-5 \, V}) = 10 \, V} \] Step 2: Calculating the work done. The work done \( W \) on a charge \( q \) moving through a potential difference \( V \) is given by: \[ W = qV \] For a proton, the charge \( q = 1.602 \times 10^{-19} \, Coulombs} \). Thus, the work done on the proton when moving from P1 to P2 is: \[ W = 1.602 \times 10^{-19} \, C} \times 10 \, V} = 1.602 \times 10^{-18} \, J} \] Rounding off slightly, the work done matches the value given in option (B).