Asked by Martin

A test charge of 5.0x10^-6 C is moved 2.0 cm through an electric force of 6.0 x 10^-7 N .
What is the change in evergy of the test charge?
What is the change in voltage on the charge?

Help
Answered by Damon
I need to know the signs. If you are pushing the test charge up then the potential energy goes up.
Work = force * distance = increase in energy

= 6*10^-7 * 2* 10^-2 = 12 * 10^-9 = 1.2 * 10^-8 Joules

Voltage change = energy change / charge
= 1.2 * 10^-8 / 5 * 10*-6 = .24 *10^-2
= 2.4 * 10^-3 volts
Answered by Martin
Hello damon... I'm not sure what happened there... what happened to the 5.0 x 10^-6 Could you show me again, as I don't quite understand it.
thanks
martin
Answered by Damon
You did not need the 5*10^-6 for the first part of the question. You were given the force and work = change in energy = force times distance

You do need it in the second part where you divide the energy by the charge to get voltage (which is potential energy per unit charge)

1.2 / 5 = .24 and 10^-8/10^-6 = 10^-2
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