So the electric field between two parallel plates is given by $E = V/d.$ How do you derive this?


2 Answers 2


Recall that the potential difference between two points $a$ and $b$ is given by $$\Delta V=-\int_{a}^{b} \vec{E}\cdot d\vec{\ell}.$$ Consider evaluating this integral for two paralell plates, i.e. the point $a$ is in one plate and the point $b$ is in the other plate. Then, we know that the electric field between paralell plates (assuming they are very close together) is of the form $$\vec{E}=E\hat{x},$$ where $\hat{x}$ is a unit vector perpendicular to any of the plates. Now, because the path integral that I quoted for the potential difference is path independent, I can take $d\vec{\ell}=d\vec{x}=dx\hat{x}$. Then: $$\Delta V=-\int_{a}^{b} E\hat{x}\cdot dx\hat{x}.=-\int_{a}^{b}Edx=E(a-b).$$ In your notation, $\Delta V=V$ and $(a-b)=d$ (the sign is just a matter of the use), so, translating the above result we have $$V=Ed \Longrightarrow E=\frac{V}{d}.$$


The second more complex possibility (but without integrals) is using the expression for capacitor

$$Q = V C$$

Since the total charge is

$$Q = \sigma A$$

and electrical field of one charged plate is

$$E' = \frac{\sigma}{2 \epsilon_0}$$

noting that there are two plates with opposite fields you get

$$E = \frac{\sigma}{\epsilon_0}$$

Combining those with expression for parallel plate capacitance

$$C = \frac{\epsilon_0 A}{l}$$

you get your expression.

But the usual derivation goes in the opposite direction ;-)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.