Question

The energy of a certain charged capacitor is 7 J. What is the new energy stored in that capacitor if its charge is decreased to 1 / 4 of its original value (imagine allowing some of the charge to move through a resistor)? Remember that the capacitance, C , that relates Q and V is unchanged.

Enew=?

Answer 1

Answer:

The new energy is 1/16 of the original energy

Explanation:

A capacitor is an electric device which is able to store charge when it is connected to a power supply.

The energy stored in a capacitor is given by the equation

$E=\frac{Q}{2C}$

where:

Q is the charge stored on the capacitor

C is the capacitance of the capacitor

For the capacitor in this problem, initially we have

E = 7 J (energy) when the charge stored is Q and the capacitance is C

Later, the charge is decreased to 1/4 of its original value, so the new charge is

$Q'=\frac{1}{4}Q$

Since the capacitance remains the same, the new energy is

$E'=\frac{(\frac{1}{4}Q)^2}{2C}=\frac{1}{16}\frac{Q^2}{2C}=\frac{1}{16}E$

Therefore, the new energy is 1/16 of the original energy.