Question

The work done by an external force to move a particle of charge -2.20 HC at constant speed from point A to point B is 6.40x10-3J. (a) What is the difference between the electric potential energy of the charge at the two points? (b) What is the difference in the kinetic energy of the charge at the two points? (c) Determine the potential difference between the two points. (d) State which point is at the higher potential.

Answer 1

(a) The electrical potential difference between the two points is equivalent to the amount of work done. (Since the charge is moved at constant speed, there is no energy used for changing speed, ie. No Kinetic energy change). Therefore the electrical potential energy difference is,

$\Delta U = 6.40 \times 10^{-3} \, J$

(b) Since the charge is moved at constant velocity, there will be no change in Kinetic Energy of the charge in the two points.

$\Delta KE = 0$

(c) The potential difference between the two points can be calculated as,

$E = q \Delta V \Rightarrow \Delta V = \frac{E}{q}$

$\Rightarrow \Delta V = \frac{6.40 \times 10^{-3}}{-2.2 \times 10^{-6}} = -2909.09 \, V$

(d) Since the potential difference is negative, the final point has lesser potential as compared to the initial point where the charge was. ie. Point A has higher potential than B.