My name is Alexander FufaeV and here I will explain the following topic:

How to Derive Potential, E-field, Charge and Capacitance of a Plate Capacitor

Table of contents

Plattenkondensator mit einer Spannung $U$ zwischen den beiden Elektroden. Abstand der Elektroden ist $d$ und die Fläche einer Elektrode ist $A$. Dazwischen ist ein Dielektrikum $\varepsilon_{\text r}$.

Here we want to derive electric potential, electric field and capacitance.

Electric potential in plate capacitor

In the following, the electric potential $\varphi$ between two capacitor plates is derived. The plate area is $A$ and the plates are at a distance $d$ from each other.

For the derivation of the potential, Poisson's equation is used:

$$ \begin{align} \nabla^2 \, \varphi ~=~ \frac{\partial^2 \varphi}{\partial x^2}+\frac{\partial^2 \varphi}{\partial y^2 }+\frac{\partial^2 \varphi}{\partial z^2} ~=~ - \frac{\rho}{\varepsilon_0} \end{align} $$

Since there are no charges between the plates, the charge density $\rho$ (charge per volume) inside is zero. The Poisson equation simplifies to the Laplace equation:

$$ \begin{align} \nabla^2 \, \varphi ~=~ \frac{\partial^2 \varphi}{\partial x^2}+\frac{\partial^2 \varphi}{\partial y^2 }+\frac{\partial^2 \varphi}{\partial z^2} ~=~ 0 \end{align} $$

The three-dimensional Laplace equation 2 can be reduced to the one-dimensional case because the plates are homogeneously charged and the potential $\varphi$ is thus independent of the $y$ and $z$ coordinates:

$$ \begin{align} \frac{\partial^2 \varphi}{\partial x^2} ~=~ 0 \end{align} $$

Thus, to find the electrostatic potential $\varphi$ inside the plate capacitor, the differential equation 3 must be solved. However, this is quite simple, because the second spatial derivative which is zero corresponds to a linear function:

$$ \begin{align} \varphi(x) = a \, x ~+~ b \end{align} $$

Here $a$ and $b$ are constants. That this form of $\varphi$ must be correct can easily be checked by differentiating twice with respect to $x$. It yields zero, as required by Laplace's equation 3.

Now the slope $a$ and the y-axis intercept $b$ of the function $\varphi(x)$ must be determined, because they are still unknown. For this purpose the boundary conditions of the problem are used.

Boundary condition #1: The first capacitor plate is placed at $x=0$ and has the constant potential $\varphi_1$ there.
Boundary condition #2: The second capacitor plate is placed at $x=d$ and has the constant potential $\varphi_2$ there.

Now the first boundary condition is substituted into 4:

$$ \begin{align} \varphi_1 = b \end{align} $$

Thus $b$ is now determined. Here $b$ represents the potential at the first plate. Now the second boundary condition must be used. Substitute it into 4:

$$ \begin{align} \varphi_2 = a \, d + b \end{align} $$

In Eq. 6 $b$ occurs. But we have already determined $b$ in Eq. 5:

$$ \begin{align} \varphi_2 = a \, d + \varphi_1 \end{align} $$

But the potentials $\varphi_1$ and $\varphi_2$ are not known either. What is known, however, is the applied voltage $U$ between the capacitor plates! It is given by the potential difference:

$$ \begin{align} U = \varphi_1 - \varphi_2 \end{align} $$

Rearrange 7 for $\varphi_1 - \varphi_2$ and insert voltage 8:

$$ \begin{align} -U = a \, d \end{align} $$

The voltage $U$ and the distance $d$ of the plates are known and therefore the slope is also known:

$$ \begin{align} a = -\frac{U}{d} \end{align} $$

Electric potential $\varphi$ of a plate capacitor. One electrode was placed on the coordinate $x=0$ and the other electrode on the coordinate $x=d$.

Now the determined constants 5and 10 have to be inserted into the potential equation 4 to get the potential inside the capacitor:

$$ \begin{align} \varphi_x = - \frac{U}{d} \, x ~+~ \varphi_1 \end{align} $$

How does the potential behave in the plate capacitor?

The potential in the plate capacitor decreases linearly from the positively charged to the negatively charged plate.

Electric field in plate capacitor

To express the electric field using the known voltage $U$, the spatial derivative of the potential (gradient equation) is used (in the one-dimensional case):

$$ \begin{align} E = -\frac{\partial \varphi}{\partial x} \end{align} $$

Differentiating the previously determined potential 11 gives the electric field inside the capacitor:

$$ \begin{align} \class{purple}{E} ~=~ \frac{U}{d} \end{align} $$

How does the E-field change when the plate spacing is reduced?

The smaller the plate spacing with the voltage held constant, the larger the electric field between the plates.

Capacitance of the plate capacitor

In the following, the capacitance $C$ of the plate capacitor is derived, which tells us how good the plate capacitor can 'store' the electric charge.

The electric field $E$ of a charged plate is given by:

$$ \begin{align} E = \frac{\sigma}{\varepsilon_0} \end{align} $$

Here $\sigma = Q/A$ represents the surface charge density of the charged plate. So $ \sigma $ is charge $Q$ per plate area $A$:

$$ \begin{align} E = \frac{1}{\varepsilon_0} \, \frac{Q}{A} \end{align} $$

The electric field is proportional to the charge according to 15. And since the electric field is also proportional to the voltage according to 13, the voltage is proportional to the charge. The constant of proportionality $C$ is called the electric capacitance:

$$ \begin{align} Q = C \, U \end{align} $$

In the case of a plate capacitor, the charge $Q$ on the plate is unknown; the capacitance $C$ is also unknown. Only the voltage $U$ is given by a voltage source and is thus known. So the goal is to figure out the charge of the capacitor in order to be able to calculate the remaining unknown, namely the capacitance $C$.

Equate 15 with 13 to have an equation for $Q$ that contains only known quantities. Then rearrange the resulting equation for $Q$:

$$ \begin{align} Q = \varepsilon_0 \, \frac{A}{d} \, U \end{align} $$

Now just insert 17 into 16 and rearrange for the capacitance:

$$ \begin{align} C = \varepsilon_0 \, \frac{A}{d} \end{align} $$

How can I increase the capacitance of a plate capacitor?

To increase the capacitance, increase the area of the capacitor plates and decrease their distance to each other.