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 = 2.8), how much energy could this modified capacitor store when connected to the same battery?

Answer 1

Answer:

$5.6\times 10^{-4}\ J$

Explanation:

d = Distance between capacitors

V = Voltage

k = Dielectric

A = Area

$\epsilon_0$ = Permittivity of free space

Energy the capacitor stores

$E=\frac{1}{2}CV^2=1\times 10^{-4}\ J$

Capacitance is given by

$C=\frac{k\epsilon_0A}{d}\\\Rightarrow C=\frac{\epsilon_0A}{d}$

The new capacitance will be

$C'=\frac{k\epsilon_0A}{\frac{d}{2}}\\\Rightarrow C'=\frac{2.8\epsilon_0A}{\frac{d}{2}}\\\Rightarrow C'=\frac{2.8\epsilon_0A\times 2}{d}\\\Rightarrow C'=\frac{5.6\epsilon_0A}{d}\\\Rightarrow C'=5.6C$

New energy will be

$E'=\frac{1}{2}CV^2\\\Rightarrow E'=\frac{1}{2}5.6CV^2\\\Rightarrow E'=5.6\frac{1}{2}CV^2\\\Rightarrow E'=5.6E\\\Rightarrow E'=5.6\times 1\times 10^{-4}\\\Rightarrow E'=5.6\times 10^{-4}\ J$

The energy the modified capacitor store when connected to the same battery is $5.6\times 10^{-4}\ J$