Posted by Martin on Saturday, May 19, 2012 at 2:49pm.
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?
Physics 30 - Damon, Saturday, May 19, 2012 at 3:48pm
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
Physics 30 - Martin, Saturday, May 19, 2012 at 4:08pm
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.
Physics 30 - Damon, Saturday, May 19, 2012 at 4:41pm
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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