An empty capacitor is capable of storing 1.0 × 10-4 J of energy when connected to a certain battery. If the distance between the plates is halved and then filled with a dielectric (κ = 3.3),

Question

An empty capacitor is capable of storing 1.0 × 10^-4 J of energy when connected to a certain battery. If the distance between the plates is halved and then filled with a dielectric (κ = 3.3), how much energy could this modified capacitor store when connected to the same battery?

Answer 1

The Voltage will remain constant and the capacitance will change. The energy is 1/2CV². Therefore, if we find the new Capacitance, C', then E'/E = C'/C because the energy is proportional to the capacitance for the same voltage.

The capacitance is related to the κ (dielectric constant), A (Area) and d (distance between the plates by the formula: C = κε₀A/d

C'/C = (κ'/κ)(d/d') = 3.3*2 = 6.6 because the capacitance is proportional to the dielectric constant and inverse with the distance between the plates. (A kept constant)

Therefore E'/E = 6.6 or E' = 6.6 x 10^-4 J