Question

The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started from rest and had 1.50×10−4 J of kinetic energy when it reached point b, what must be the potential difference between a and b?

Answer 1

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

$W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J$

The change in electrostatic potential energy $\Delta U$, of one point charge q is defined as the product of the charge and the potential difference.

$\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V$