Figure 21.70 shows a close-up of the central region of a capacitor made of two large metal plates of area , very close together and charged equally and oppositely. There are +Q and -Q on the inner surfaces of the plates and small amounts of charge +q and -q on the outer surfaces.

(a). Knowing Q, determine E: Consider a Gaussian surface in the shape of a shoe box, with one end of area A_box in the interior of the left plate and the other end in the air gap (surface 1 in the diagram). Using only the fact that the electric field is expected to be horizontal everywhere in this region, use Gauss’s law to determine the magnitude of the field in the air gap. Check that your result agrees with our earlier calculations in the Chapter 15. (Be sure to consider the flux on all faces of your Gaussian box.)

Consider a Gaussian surface in the shape of a shoe box, with one end of area $A_{\text{box}}$in the interior of the left plate and the other end in the air gap.

Gauss law relates the flux through a surface and the net charge by the surface. The electric field$\overrightarrow{E}$ in the region is given by the expression $\oint \overrightarrow{E}·\hat{n}dA=\frac{q_{\text{inside}}}{{\epsilon }_0}$.

Here the normal vector to the surface is $\hat{n}$, the area of the Gaussian surface is dA. If the medium in between the conductors in air, therefore, the permittivity of free space is ${\epsilon }_0$.

The charge +Q is distributed on the metal plate of area A. The charge inside the shoebox of the area$A_{\text{box}}$is given by:

$q_{\text{inside}}=\frac{Q}{A}\times A_{box}\left(2\right)$

The flux emerges normally from the surface of the plates of the capacitor. Therefore, the angle $\theta$between $\overrightarrow{E}$and$\hat{n}$is zero for the faces parallel to the plate of the capacitor and$90°$to the faces perpendicular to the plate. Therefore, for the perpendicular surfaces,

$$\begin{gathered} \backslash \overrightarrow{E}·\hat{n}dA=EdA\cos \theta \\ \overrightarrow{E}·\hat{n}dA=EdA\cos 90° \\ \overrightarrow{E}·\hat{n}dA=0 \end{gathered}$$

But for the parallel plates:

$$\begin{gathered} \overrightarrow{E}·\hat{n}dA=EdA\cos \theta \\ \overrightarrow{E}·\hat{n}dA=EdA \end{gathered}$$

One of the parallel surface lies inside the plate. The field in the interior of the conducting plate is zero, hence no flux emerges through the surface in the plate.

The flux therefore emerges only through the surface parallel to the plate and lying outside the plate.

As the electric field is constant over the area considered,

$\oint EdA=E\oint dA$

The area integral is given by:

$\oint dA=A_{\text{box}}\left(3\right)$

Now using the equation (2) and (3) in equation (1) as:

$$\begin{gathered} E{A}_{\text{box}}=\frac{Q}{A}\times \frac{A_{\text{box}}}{{\epsilon }_0} \\ E=\frac{Q}{A{\epsilon }_0} \end{gathered}$$

Therefore, the electric field in region (1) will be $E=\frac{Q}{A{\epsilon }_0}$.