Question
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
What is the change in evergy of the test charge?
What is the change in voltage on the charge?
Help
Answers
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
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
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
thanks
martin
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
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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