Question

A parallel-plate capacitor has an area of 4.0 cm^2, and the plates are separated by 1.0 mm.
a. What is the capacitance?
b. How much charge does this capacitor store when connected to a 9.0 V battery?

Answer 1

Answer:

$C=3.54\cdot 10^{-12}\ F$

$Q=3.186\cdot 10^{-11}\ c$

Explanation:

Capacitance of a parallel-plate capacitor

The parallel plate capacitor consists of two identical conducting plates, each having a surface area A, separated by a distance d with no material between the plates.

The capacitance of the capacitor can be calculated as follows:

$\displaystyle C=\epsilon_o\frac{A}{d}$

Where $\epsilon_o$ is the permittivity of free space. Its numerical value in SI units is:

$\epsilon_o=8.85 \cdot 10^{-12}\ F/m$

a.

The capacitor of the question has an area of A=4 cm^2 and the plates are separated by d=1 mm. Both magnitudes must be converted to SI:

$\displaystyle A=4\ cm^2\left(\frac{1\ m}{100\ cm}\right)^2=4\cdot 10^{-4}\ m^2$

$\displaystyle d=1\ mm\frac{1\ m}{1000\ mm}=1\cdot 10^{-3}\ m$

Now calculate the capacitance:

$\displaystyle C=8.85 \cdot 10^{-12}\frac{4\cdot 10^{-4}\ m^2}{1\cdot 10^{-3}\ m}$

$C=3.54\cdot 10^{-12}\ F$

b.

The charge Q stored in any capacitor C when a voltage V is applied is given by the equation:

Q = CV

Entering the known values into this equation gives:

$Q = 3.54\cdot 10^{-12}\cdot 9$

$Q=3.186\cdot 10^{-11}\ c$