Question

A 6.00-mu F capacitor is connected to a 12.0-V battery. How much energy is stored in the capacitor? to the product of the capacitance of the capacitor times the potential difference across the capacitor. Why isn't this equal to the stored energy? Had the capacitor been connected to a 6.00-V battery, how much energy would have been stored? How does the energy stored in a capacitor depend on the potential difference across the capacitor?

Answer 1

Answer:

Explanation:

C = 6 x 10^-6 F, V = 12 V,

Energy stored in the capacitor

E = 1/2 CV^2 = 0.5 x 6 x 10^-6 x 12 x 12 = 4.32 x 10^-4 J

Now the voltage is 6 V, So let energy be E'

E' = 1/2 CV^2 = 0.5 x 6 x 10^-6 x 6 x 6 = 1.08 x 10^-4 J

The energy stored in a capacitor is proportional to the square of teh potential difference applied across the capacitor.