If you increase the charge on a parallel-plate capacitor from 3 µC to 9 µC and increase the plate separation from 1 mm to 3 mm,the energy stored in the capacitor changes by a factor of
A)27
B)9
C)3
D)8
E)1/3

Question

Quizplus · Accepted Answer

The energy stored in a parallel-plate capacitor is given by the equation:$$E = \frac{1}{2}CV^2$$where:* E is the energy stored in joules (J)* C is the capacitance in farads (F)* V is the voltage across the capacitor in volts (V)If we increase the charge on the capacitor from 3 µC to 9 µC, the voltage across the capacitor will also increase by a factor of 3. This is because the capacitance of the capacitor remains constant, and the voltage is directly proportional to the charge.If we increase the plate separation from 1 mm to 3 mm, the capacitance of the capacitor will decrease by a factor of 3. This is because the capacitance is inversely proportional to the plate separation.Therefore, the energy stored in the capacitor will change by a factor of:$$E = \frac{1}{2}CV^2 = \frac{1}{2}C(3V)^2 = 27\frac{1}{2}CV^2$$So, the energy stored in the capacitor changes by a factor of 27.

If You Increase the Charge on a Parallel-Plate Capacitor from 3