# Proving electric field constant between two charged infinite parallel plates

It is known that the electric field intensity between two infinitely long charged parallel plates is constant.

I had read that one explanation is that if a test charge is placed between the plates, both the force by the positive plate on the test charge and the negative plate test charge acts in the same direction, and the sum of these two forces is the same at any point between the plates. How do you show mathematically that this is true?

Edit: I noticed the answers on proving this using Gauss law are excellent. Is there a way to prove this using Coulomb's law? Which means using $\vec F =\frac{q_1q_2}{4 \pi \epsilon_0 r^2}$ to show that the resultant force is the same for any value of $r$?

• This is a question which is often answered in textbooks and the Internet using Gauss's law. Here is a link in which Coulomb's law (and Gauss's law) are used. web.mit.edu/sahughes/www/8.022/lec02.pdf – Farcher Aug 15 '18 at 5:57

Hint :

Suppose that a plane $\;\mathcal P_1\;$ parallel to the $\;xy\;$ plane at $\;z=0\;$ has a uniform surface charge density $\;\sigma_1$. Then
$$\mathbf{E}_1\left(x,y,z\right)= \left. \begin{cases} \boldsymbol{+}\dfrac{\sigma_1}{2\epsilon_0}\mathbf{k},\, \text{for } z>0 \vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \boldsymbol{-}\dfrac{\sigma_1}{2\epsilon_0}\mathbf{k},\, \text{for } z<0 \end{cases}\right\} \tag{01}\label{eq01}$$ where $\;\mathbf{k}\;$ the unit vector on axis $\;z$.

Now, take an other plane $\;\mathcal P_2\;$ parallel to the $\;xy\;$ plane at $\;z=h\;$ with a uniform surface charge density $\;\sigma_2\;$ producing its own $\;\mathbf{E}_2\;$ and find $\;\mathbf{E}\left(x,y,z\right)=\mathbf{E}_1\left(x,y,z\right)+\mathbf{E}_2\left(x,y,z\right)$.

Proof of equation \eqref{eq01} using Coulomb's Law only

In Figure-01 a test charge $\;q\;$ is at a height $\;h\;$ over a plane $\;\mathcal P\;$ with uniform surface charge density $\;\sigma$. Let a ring of finite radius $\;R\;$ and of infinitesimal width $\;\mathrm dR$. If we take a piece of the ring of infinitesimal length $\;\delta\ell\;$ then the infinitesimal area $\;\mathrm dR\cdot\delta\ell\;$ carries charge $\;\sigma\cdot\mathrm dR\cdot\delta\ell\;$ and exerts on the charge $\;q\;$ a force
$$\delta\mathbf{f}=\dfrac{q\cdot\left(\sigma\cdot\mathrm dR\cdot\delta\ell\right)}{4\pi\epsilon_{0}r^2}\dfrac{\mathbf{r}}{\;r\;} \tag{02}\label{eq02}$$ This force has components normal and parallel to the plane $\;\mathcal P\;$ $$\delta\mathbf{f}=\delta \mathbf{f}_{\boldsymbol z}\boldsymbol{+}\delta \mathbf{f}_{\boldsymbol{\rho}} \tag{03}\label{eq03}$$ If we integrate with respect to $\;\ell\;$ we have $$\underbrace{\oint_\ell\delta\mathbf{f}}_{\mathrm d \bf F}=\underbrace{\oint_\ell\delta \mathbf{f}_{\boldsymbol z}}_{\mathrm d \mathbf{F}_{\!\boldsymbol z}}\boldsymbol{+}\underbrace{\oint_\ell\delta \bf f_{\boldsymbol{\rho}}}_{\mathrm d \bf F_{\! \boldsymbol{ \rho}}} \tag{04}\label{eq04}$$ or $$\mathrm d \mathbf{F}=\mathrm d \mathbf{F}_{\!\boldsymbol z}\boldsymbol{+}\mathrm d \mathbf{F}_{\! \boldsymbol{ \rho}} \tag{05}\label{eq05}$$ Because of the rotational symmetry with respect to the vertical $\;z-$axis the components $\;\delta \bf f_{\boldsymbol{\rho}}\;$ cancel out so $\;\mathrm d \bf F_{\!\boldsymbol{ \rho}}=\boldsymbol{0}\;$ and $$\mathrm d \mathbf{F}=\mathrm d \mathbf{F}_{\!\boldsymbol z}=\dfrac{q\sigma}{2\epsilon_{0}}\dfrac{R\cos\theta\mathrm dR}{r^2}\mathbf{k} \tag{06}\label{eq06}$$ Now from the geometry of this configuration \begin{align} R &=h\tan\theta \tag{07.1}\label{eq07.1}\\ \mathrm dR &=\dfrac{h}{\cos^2\theta}\mathrm d\theta \tag{07.2}\label{eq07.2}\\ r^2 &=\dfrac{h^2}{\cos^2\theta} \tag{07.3}\label{eq07.3} \end{align} and replacing in \eqref{eq06} $$\mathrm d \mathbf{F}=\dfrac{q\sigma}{2\epsilon_{0}}\sin\theta\,\mathrm d\theta\,\mathbf{k} \tag{08}\label{eq08}$$ Integrating $$\mathbf{F}=\dfrac{q\sigma}{2\epsilon_{0}}\left(\int\limits_{0}^{\pi/2}\sin\theta\,\mathrm d\theta\right)\mathbf{k}=\dfrac{q\sigma}{2\epsilon_{0}}\Bigl[-\cos\theta\Bigr]_{0}^{\pi/2}\mathbf{k}=\dfrac{q\sigma}{2\epsilon_{0}}\mathbf{k} \tag{09}\label{eq09}$$ so $$\mathbf{E}=\dfrac{\mathbf{F}}{q}= \dfrac{\sigma}{2\epsilon_{0}}\,\mathbf{k} \tag{10}\label{eq10}$$ a result independent of the position coordinates of the test charge $\;q$.

In Figure-02 we see the force exerted on $\;q\;$ by a ring of infinitesimal width, equation \eqref{eq08}.

In order to show that this is indeed the case you will need to use so called Gauss law. It relates flux of electric field through closed surface with charge contained inside: $$\oint \vec{E}d\vec{S} = \frac{Q}{\epsilon_{0}}$$

In general it is quite difficult to evaluate left hand side since orientation of area element $d\vec{S}$ of closed surface wrt $\vec{E}$ may vary from point to point. In some cases however it is possible to define so called Gaussian surface. It's given by the orienting $d\vec{S}$ perpendicular to $\vec{E}$(or collinear to it). For instance if you want to apply Gauss law to evaluate electric field of point charge you should pick sphere with center placed at the position of point charge.

Now we need to understand how to pick Gaussian surface for infinite plane. It is useful to start from following toy model. We place point charges uniformly along the surface. Due to symmetry of construction one can observe that electric field of nearest charges will cancel each other in all directions apart from the one exactly perpendicular to the charged surface.(Blue lines will stay uncanceled. Vertical parts of purple lines will cancel corresponding parts of nearest green ones.

You can see that in order for this cancellation to take place you need to have

1) uniformly charged plane

2) plane to be large(there won't be cancelation near the boundary of plane)

Now if we go to continuous limit we arrive to the conclusion that electric field is perpendicular to the surface of plane.(Note that the same thing will take place from the opposite side of the plane). Since we know orientation of electric field we can easily define corresponding Gaussian surface:

Now Gauss law will give us:

$$E\cdot 2 A = \frac{Q}{\epsilon_0} = \frac{\sigma A}{\epsilon_0}$$

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

here $\sigma$ is surface charge density, $A$ - surface area of top cylinder.

So you don't really need to have two planes in order to have constant electric field.

If you take two identical infinite planes with opposite charges electric field inside will double and will cancel outside the planes.