voltage question?

Question

What is the voltage change when
a) an electric field does 12J of work on .0001C charge?
b) the same electric field does 24J of work on a .0002C?

Answer 1

Tlbs101's answer isn't quite right, because he is assuming a capacitor that is being charged, which isn't the situation you describe. Your question is about the work that an E field does to some charge, as in an electron gun, which is inside every CRT.

The correct equation is Energy = delta-Voltage x Charge.

It is the delta-voltage that causes the electric field.

So, delta-voltage = Energy / Charge.

I won't do the math for you, but I will tell you that the answer to a) and b) are the same.

Answer 2

Capacitance [Farads] = Charge [Coulombs] / Voltage [volts]

Energy [Joules] = 1/2 * Capacitance * Voltage^2
therefore E = 1/2 * Charge * voltage

12 Joule = 1/2 * 0.0001 Coulombs * V volts
V = 240,000 volts

that checks... the capacitance is 416.66 pF. and in order to have 12 J on 416 pF you need something like a quarter million volts.

You can probably figure out b) for yourself, now.

EDIT: Robert T is correct. For an open-field E-field (not that found in a capacitor, my answer is off by a factor of 2 (as in 2 plates of a capacitor, versus 1 point of charge in the problem to infinity).

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