Electric Field Between Two Parallel Plates

The Electric Field Strength between Two Parallel Plates

The strength of the electric field \(E\) that exists between the plates is related to the potential difference between the plates \(V\) as well as the separation between the plates \(r\) by the equation \[E=\frac{V}{r}.\] The SI unit of measurement for electric field strength is \(\mathrm{V\,m^{-1}}.\) We can assume that over the distance \(r\), the potential difference \(V\) will change at a constant rate, and so we can write this equation as follows, \[E=\frac{\Delta V}{\Delta r},\] where \(\Delta V\) is a small change in the potential difference over a small distance \(\Delta r.\)

The electric field should also depend on the physical structure of the plates themselves. There is yet another equation that gives us the electric field strength between two parallel plates that depends on the charge on one of the plates \(Q\) and the surface area of a plate \(A.\) This equation is \[E=\frac{Q}{\varepsilon_{0}A},\] where \(\varepsilon_{0}\) is a constant known as the permittivity of free space which indicates how well electric fields can pass through vacuum. The value of this constant is \(8.85 \times 10^{-12}\,\mathrm{m^{-3}\,kg^{-1}\,s^{4}\,A^{2}}.\) This is equivalent to \[\varepsilon_{0}=8.85 \times 10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}}.\] The figure below provides a visualization of the parallel plates with both their areas and magnitudes of charge being equal.

The Electric Field Strength between Two Parallel Plates: Derivation

If we look back at the scenario from the first figure concerning the charged particle in the region between the plates, we can derive the equation for the electric field that we have stated above. The work done to move the charge from one plate to another is the product of the force \(F\) and the distance moved by the charge in the direction of the force \(r\), \[W=Fr.\] We also know that potential difference is defined as the work done per unit charge, \[\begin{align}V&=\frac{W}{q} \\\Rightarrow W&=Vq.\end{align}\] By equating the two expressions for the work done we get \[\begin{align}Fr&=Vq\\\Rightarrow\frac{F}{q}&=\frac{V}{r}\\\Rightarrow E&=\frac{V}{r},\end{align}\] since the electric field strength \(E\) is defined to be the force per unit charge \(F/q.\) This is the same equation as the one stated above.

The Electric Field between Two Parallel Plates of the Same Charge

We want to now imagine what would happen if the charge on both of the plates were equal. Let's consider the scenario in our very first image above; if the charge from both plates were equal in magnitude and sign, there would be no potential difference between the plates. That is, the work done per unit charge would be zero, and the particle would not move from one plate to the other. Mathematically, for the electric field strength, we get \[\begin{align}E&=\frac{V}{r}\\&=\frac{0}{r}\\&=0\,\mathrm{V\,m^{-1}}.\end{align}\] So the electric field strength in the region between the plates would also be zero.

The Electric Field between Two Parallel Plates: Calculation

We need to now test our new knowledge of the electric field from two parallel plates in the following examples. We will first solve for the field strength given the plate separation and potential difference.

Let's now try to solve for the electric field strength given the charge on one of the plates and the surface area.

So the electric field strength can be calculated if we know the potential difference and separation between the plates; or if we know the charge and area of a plate.

Motion in a Uniform Electric Field vs Projectile Motion

If a charged particle enters a uniform electric field, it experiences an electric force that is the same on it at all points in the field. This is analogous to the scenario in which a particle with mass enters a uniform gravitational field; it will feel the same gravitational force at all points in the field. The gravitational field near the Earth's surface is approximated to be uniform since it is relatively unchanged at distances close to the Earth's surface.

This means that the equations that govern projectile motion for massive objects would have a similar form to that of charged particles moving in a uniform electric field. Fig. 4 below shows a positively charged particle moving at some angle relative to the surface of the plates. It will undergo parabolic motion that is similar to projectile motion, but the force on the charge is electrostatic in nature and not gravitational.

Fig. 4 - A charged particle moving the uniform field between parallel plates would undergo the same motion that a massive particle would on the Earth.

It enters the region containing the electric field \(E\) with initial velocity \(u\), which can be broken down into its horizontal and vertical components, \(u_x\) and \(u_y.\) It moves along a parabolic trajectory as there is an electric force that acts on it throughout its motion. It then leaves the field with velocity \(v\) in a straight line, since the force no longer acts on it outside the region between the plates.