- #1

- 191

- 0

## Homework Statement

The work done by an external force to move a -8.50 mu C charge from point a to point b is 1.90×10^-3 J.

If the charge was started from rest and had 4.86×10^-4 J of kinetic energy when it reached point b, what must be the potential difference between a and b?

## Homework Equations

V = PE / q

Vba = - W/q

## The Attempt at a Solution

If the charge started from rest, all the energy it had was potential energy which got converted to kinetic energy at point b.

V = 4.86 x 10^-4 / -8.50 x 10^-6 C

V at point a = -57.176

Vb - (-51.176 V) = -1.90 X 10^-3 J / -8.5 X 10^-6 C

Vb + 51.716 V = 223.5 V

Vb = 165.76 (166 V)

I use MasteringPhysics and it told me the answer should have been negative but I don't understand why. My book states "When the electric force does positive work on a charge, the kinetic energy increases and the potential energy decreases. The difference in potential energy, PEb - PEb, is equal to the negative of the work, Wba, done by electric field to move the charge from a to b."