# Calculating the electric field strength in parallell plate capacitor

Two parallell quadratic plates with side $$\ell$$, distance between $$d$$ and between the plates there is a material with relative permittivity $$\varepsilon_r$$ and conductivity $$\sigma.$$ By applying a potential difference of $$120$$ V over the plates a current will flow. (A) Determine the electric field strength $$|\vec{E}|$$ between the capacitor plates.

What annoys me the most is that I've been given several nice formulas to calculate the electric field, for example

$$\varepsilon\oint_S\vec{E}\cdot \ d\vec{S} = Q_{\text{enclosed}} \quad \text{Gauss law} \tag 1$$ $$\vec{E}(\vec{R_2})=\frac{Q}{4\pi\varepsilon}\iint_S\frac{\rho_s(\vec{R_1})}{R_{12}^2} \ \widehat{R}_{12} \ dS \quad \text{Brute force} \tag 2$$

$$\vec{E}=-\nabla V \quad \text{If potential is given} \tag3$$

These are the only formulas I'm given, only formulas the book uses. However on the above particular question from an old exam, I'm suddenly expected to use the petty $$|\vec{E}|=V/d.$$ I understand this formula works for uniform fields, but how am I supposed to derive it using one of the 3 formulas given on the exam formula sheet?

If $$\ell \gg d$$, then you can use Gauss' Law to show the field between the plates (far from the edges) is uniform.
$$V=-\int \vec{E}\cdot d\vec{\ell},$$
to get $$\left|\vec{E}\right|=V/d$$.