Question

Problem 2. Stored energy on a capocitor Two parallel plates are charged up by a battery and then the battery is removed. Then we move the plates further apart. (a) Draw the capacitor before and after the plates are moved apart and label the most salient attributes. (b) Does the (1) charge, (2) charged density. (3) electric field, (4) voltage, (5) stored energy, and (6) capacitance increase, decrease, or stay the same? Explain in words without using equations.
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Answer #1

a) 3 2

b) The capacitance is charged to a charge Q then disconnected from battery.

1) Charge: As the distance between the plates is increased, charge Q on the plates stays the same.

2) Charge density: Charge density is charge per unit area. Area of the plates remains the same and charge remains the same, so charge density \sigma also stay the same.

3) Electric field: Electric field of a parallel plate capacitor depends on charge density , so electric field E also stay the same.

4)Voltage: Voltage is work done on a unit positive charge in moving it from positive plate to negative plate. As the plate seperation increases, work done also increases. So voltage V between the plates also increases .

5)Stored energy: Work done in moving the capacitor plates against the electrostatic attraction is stored as energy in the capacitor. As the distance between the plates increases, potential between the plates increses and so the stored energy U also increases.

6) Capacitance: As the charge remains the same and potential between the plates increases, the capcitance C of the parallel plates decreases.

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