Particle Deflection in an Electric Field

Electron Deflection

If we have two parallel plates with a voltage difference \(V\) then there will be a uniform electric field between the plates. If the separation of the plates is \(d\) then the field strength is

\(E=\frac{V}{d}\)

Once we know the field strength it’s easy to calculate the force on a charge \(q\), we simply multiply by the field strength.

\(F = Eq = \frac{Vq}{d}\)

A Level questions usually have an electron whizzing between the plates and ask about the force exerted on the electron by the plates, the resulting acceleration and the final velocity and direction of the electron. Let’s try solving a problem to see how this works.

Electron Deflection

Let’s say we have two plates separated by 20 millimetres with a voltage difference of 2000 volts. The length of the plates along which the electron travels is 30 millimetres. The horizontal velocity of the electron before it passes through the plates is \(1.5\times10^7 \mbox{ms}^{-1}\)

Firstly the strength of the electric field is

\(E=\frac{2000}{20\times10^{-3}}=10^5 \mbox{V m}^{-1}\)

The force on the electron

\(F = Eq = 10^5 \times 1.6\times10^{-19}=1.6\times10^{-14} \mbox{N}\)

And by Newton’s second law \(F=ma\) the acceleration of the electron toward the positive plate is

\(a = \frac{1.6\times10^{-14}}{9.11\times10^{-31}}=1.76\times10^{16} \mbox{ms}^{-2}\)

The time spent between the plates, which we need to calculate the final vertical velocity of the electron is

\(\Delta t=\frac{30\times10^{-3}}{1.5\times10^7}= 2\times10^{-9} \mbox{s}\)

In this two nanosecond period spent between the plates the vertical velocity increases by

\(\Delta v = a \Delta t = 1.76\times10^{16} \times 2\times10^{-9} = 3.51\times10^7 \mbox{ms}^{-1}\)

Electron Deflection Calculation

We can calculate the angle the velocity of the electron makes with its original direction by working out the inverse tangent of the ration of the vertical and horizontal velocities

\(\theta=\tan^{-1}\frac{v_v}{v_h}=\tan^{-1}\frac{3.51\times10^7}{1.5\times10^7}=66.9 \mbox{ degrees}\)

We can also work out the magnitude of the velocity by applying Pythagoras’ Theorem to the horizontal and vertical components of velocity

\(v=\sqrt{(3.51\times10^7)^2 + (1.5\times10^7)^2} = 3.82\times10^7 \mbox{ ms}^{-1} \)
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ramin

Ramin is a science and maths tutor. He has a first class degree and a doctorate in Physics. He has written two finance books and also teaches adults about investment through his company PensionCraft.