Final answer:
The potential difference between point A and point B is 53.8 volts when a charge of -9.1 µC is moved by external work, with the charge gaining 2.10 × 10^-4 J of kinetic energy.
Step-by-step explanation:
The student asks about the potential difference between two points when a charge is moved by an external force. To find the potential difference (Voltage), we use the concept of work done on a point charge and its relation to potential energy and kinetic energy.
The work done by the external force is equal to the change in potential energy plus the kinetic energy gained by the charge, following the Work-Energy Principle.
The potential difference (ΔV) between points A and B is computed using the equation:
ΔV = ΔU / q
where ΔU is the change in potential energy, and q is the charge. The total work W, done by the external force, is equal to the change in potential energy plus the kinetic energy (ΔK) gained:
W = ΔU + ΔK
We then rearrange to solve for the potential energy change:
ΔU = W - ΔK
Given: W = 7.00 × 10^-4 J and ΔK = 2.10 × 10^-4 J,
ΔU = 7.00 × 10^-4 J - 2.10 × 10^-4 J
ΔU = 4.90 × 10^-4 J
Now, substituting the values into the potential difference equation:
ΔV = (4.90 × 10^-4 J) / (-9.1 × 10^-6 C)
ΔV = -53.8 V
The negative sign indicates that the potential at point A is higher than at point B. Therefore, the potential difference between point A and point B is 53.8 volts.