Question

How do you find the electric field of a capacitor?

Answer 1

Step 1: Use the superposition principle for the parallel plate capacitor

For the electric field between the plates of a parallel plate capacitor, we need to combine the electric fields due to their plates using the superposition principle, Here $E_1$ and $E_2$ have the same magnitude at all points and have the same direction as well.

$\Rightarrow \overrightarrow{\mathrm{E}}=\overrightarrow{{\mathrm{E}}_1}+\overrightarrow{{\mathrm{E}}_2}\left(\mathrm{equation}1\right)$

Where $\overrightarrow{\mathrm{E}}=$electric field, $\overrightarrow{{\mathrm{E}}_1}\mathrm{and}\overrightarrow{{\mathrm{E}}_2}=$the electric field between parallel plate capacitor

Step 2: Apply Gauss law

An electric field due to a single infinite sheet of charge is:

$\Rightarrow E=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left(\mathrm{equation}2\right)$

Where $E=$electric field, $\sigma =$surface charge density, ${\epsilon }_0=$electric constant

Step 3: Find the electric field of a capacitor

Substitute the value of $equation\left(2\right)$ in $equation\left(1\right)$ we get,

$$\begin{gathered} \Rightarrow \overrightarrow{\mathrm{E}}=\overrightarrow{{\mathrm{E}}_1}+\overrightarrow{{\mathrm{E}}_2} \\ \Rightarrow \overrightarrow{\mathrm{E}}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}+\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \\ \Rightarrow \overrightarrow{\mathrm{E}}=\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_0} \end{gathered}$$

Where $\overrightarrow{E}=$electric field, $\sigma =$surface charge density, ${\epsilon }_0=$electric constant

Hence, this gives the electric field between a parallel plate capacitor.