Calculating the electric field strength in parallell plate capacitor

Question

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?

Answer 1

If $\ell \gg d$, then you can use Gauss' Law to show the field between the plates (far from the edges) is uniform.

Once you know the field is uniform (and perpendicular to the surfaces of the plates), it's easy to simplify the definition of electrostatic potential,

$$V=-\int \vec{E}\cdot d\vec{\ell},$$

to get $\left|\vec{E}\right|=V/d$.