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Physics 30

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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

  • Physics 30 - ,

    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 - ,

    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

  • Physics 30 - ,

    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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