So the electric field between two parallel plates is given by $E = V/d.$ How do you derive this?
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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 $$V=Ed \Longrightarrow E=\frac{V}{d}.$$ |
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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 ;-) |
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