Question

A 12.5μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them.a.) How much energy is stored in the capacitor before the dielectric is inserted?b.) How much energy is stored in the capacitor after the dielectric is inserted?c.) By how much did the energy change during the insertion? Did it increase or decrease?d.) Explain why inserting the dielectric (or equivalently exchanging air with the material) causes a change in the stored energy of the capacitor.

Answer 1

Explanation:

Energy stored in a capacitor is given by:

$U=\frac{Q^2}{2C}(1]$

Here, Q is the capacitor's charge and C the capacitance.

Capacitance is given by:

$C=\frac{Q}{V}$

Where V is the potential difference. Rewriting for Q and replacing in 1:

$Q=CV\\U=\frac{C^2V^2}{2C}=\frac{CV^2}{2}$

a.) $U=\frac{12.5*10^{-6}F(24V)^2}{2}=3.6*10^{-3}J$

b.) Energy stored in the capacitor with dielectric is:

$U_k=\frac{U}{k}\\U_k=\frac{3.6*10^{-3}J}{3.75}=9.6*10^{-4}J$

c.) Energy decreased in a rate of 3.75 due to the insertion of dielectric, that is the value of dielectric constant.

d.) If we introduce a dielectric, potential difference decreases, capacitance increases and charge is the same. Therefore, accord to (1) the energy decreases.