Question

A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.
a. How much energy is stored in the capacitor?
b. The separation between the plates is now increased to 10.04 mm. How much energy is stored in the capacitor now?
c. How much work is required to increase the separation of the plates from 2.51 mm to 10.04 mm?

Answer 1

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

$E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$

Here,

V = Potential difference across the capacitor plates

Q = Charge stored on the capacitor plates

At the same time capacitance can be defined as,

$C = \epsilon_0 (\frac{A}{d})$

Here,

$\epsilon_0 =$  Vacuum permittivity constant

A = Area

d = Distance

Replacing with our values we have that,

$C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})$

$C = 1.5901*10^{-10}F$

PART A) Energy stored in the capacitor is

$E = \frac{1}{2} CV^2$

$E = \frac{1}{2} (1.5901*10^{-10})(575)^2$

$E = 2.628*10^{-5}J$

PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore

$Q = CV$

$Q = (1.59*10^{-10})(575)$

$Q = 9.1425*10^{-8}C$

Now if $d = 10.04*10^{-3}m$ we have that the capacitance is

$C = \epsilon_0 (\frac{A}{d})$

$C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})$

$C = 3.9754*10^{-11}F$

Then the energy stored is

$E = \frac{1}{2} \frac{Q^2}{C}$

$E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})$

$E = 1.051*10^{-4} J$

PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore

$W = 1.051*10^{-4} J -2.628*10^{-5}J$

$W = 7.882*10^{-5} J$