• To understand the motion that charged particles undergo when in a uniform electric field
  • To practice the use of equations and calculations on topics studied so far
  • To apply the equations of motion to particles in a uniform electric field

In this section we will be merging your knowledge of the equations of motion, parabolic motion and uniform electric fields to determine how a charged particle will travel through an electric field if it already has a velocity in some given direction.

If a charged particle is initially stationary and between two parallel plates, a uniform electric field exerts a constant force and therefore a constant acceleration on the charged particle. But what would happen if a charged particle were inserted with an initial velocity between the plates perpendicular to the electric field?
The electric field is still uniform between the plates and so there will still be a constant force exerted on the particle when its is between the plates. What would happen in the above image if the particle inserted were positive? What would happen if it were negative?

The red path shows the path of a negatively charged particle and the blue path shows the path a positively charged particle would travel through. For a given electric field, the curvature depends on the mass of the particle and the magnitude and polarity of the charged particle.

For all examples such as this we assume that there will be no air resistance slowing the charged particle down. Therefore if the charged particle has an initial horizontal velocity, when it enters a vertical electric field the horizontal component of velocity will remain unchanged and so only the vertical component of velocity will change due to the electric force

Take the following example:
An electron travelling at 2 \times 10^{6} \ ms^{-1} enters perpendicularly into an electric field at the midpoint between two parallel plates. The voltage on the upper plate is +3000 \ V and the voltage on the lower plate is -3000 \ V . The separation of the plates is by 12 \ mm

Since there is no air resistance, the horizontal component of velocity will remain unchanged when the charged particle enters the electric field, it will always be u_{h} = 2 \times 10^{6} \ ms^{-1} .

The vertical component of velocity is initially u_{v} = 0 \ ms^{-1} , but when the electron enters the electric field it will experience a force upwards and therefore will accelerate up.

Determining the electric field strength:

The potential difference in this example is V = +3,000 + -3,000 = 6,000 \ V and the plate separation is 12 \ mm = 12 \times 10^{-3} \ m

E = \frac{V}{d} = \frac{6000}{12 \times 10^{-3}} = 500,000 \ Vm^{-1}

Determining the force on the electron:

Rearranging E = {F}{q} for F gives:

F = Eq = 500000 \times 1.6 \times 10^{-19} = 8.0 \times 10^{-14} \ N

Determining the acceleration of the electron:

Rearranging F = ma for a gives:

a = \frac{F}{m} = \frac{8.0 \times 10^{-14}}{9.11 \times 10^{-31}} = 8.8 \times 10^{16} ms^{-2} $ (2 s.f.)

Because the plates are parallel, this is a uniform acceleration and so is constant.

Determining how far along the upper plate from the left hand side the electron will strike the plate: (Assuming that the plates are long enough for the electron to strike it that is…)

The equations of motion are required in both the vertical and horizontal direction.

Labelling the diagram appropriately allows us to identify what components can be worked out and when:The objective is to work out x(h) , so the SUVAT is required horizontally:

s = x(h)
u = u(h) = 2 \times 10^{6} ms^{-1}
v = u(h)
a = 0 \ ms^{-2}
t = \text{unknown}

SUVAT can only be used if three variables are known and a fourth is wanted. Therefore the vertical components are required to determine time – since time is the same for both horizontal motion and vertical motion:

s = x(v) – this is half the distance between the two plates since the electrons enter midway between the two, so s = x(v) = 6 \ mm = 6 \times 10^{-3} \ m
u = 0 \ ms^{-1}
v = \text{n/a}
a = 8.8 \times 10^{16} \ ms^{-2}
t = \text{?}

Using s = ut + \frac{1}{2}at^{2} , the time can be determined:

6 \times 10^{-3} = 0 \times t + \frac{1}{2} \times 8.8 \times 10^{16} \times t^{2}
6 \times 10^{-3} = \frac{1}{2} \times 8.8 \times 10^{16} \times t^{2} 1.3636 \times 10^{-19} = t^{2}
t = 3.69 \times 10^{-10} \ s

Now time is known, it can be substituted into SUVAT horizontally:

s = x(h) = ?
u = 2 \times 10^{6} ms^{-1}
v = \text{unknown}
a = 0 \ ms^{-2}
t = 3.69 \times 10^{-10} \ s

Using s = ut + \frac{1}{2}at^{2} , the time can be determined:

x(h) = 2 \times 10^{6} \times 3.69 \times 10^{-10} + \frac{1}{2} \times 0 \times t^{2}
x(h) = 2 \times 10^{6} \times 3.69 \times 10^{-10}
x(h) = 7.38 \times 10^{-4} \ m = 0.74 \ mm (2 s.f.)

Determining the final speed of the electron before it collides with the upper plate

To determine the final speed, both the final horizontal and final vertical speeds are required.

The final horizontal speed is the same as the starting horizontal speed, u(h) = 2 \times 10^{6} \ ms{-1}

The final vertical speed can be deduced using SUVAT:
s = x(v)= 6 \ mm = 6 \times 10^{-3}
u = 0 ms^{-1}
v(v) = ?
a = 8.8 \times 10^{16} \ ms^{-2}
t = 3.69 \times 10^{-10} \ s

Using any of the equations of motion v(v) can be determined:

v^{2} = u^{2} + 2as
v(v)^{2} = 0 + 2 \times 8.8 \times 10^{16} \times 6 \times 10^{-3}
v(v)^{2} = 1.056 \times 10^{15}
v(v) = 3.25 \ times 10^{7} \ ms^{-1}

We now know the horizontal and vertical components of the final velocity:

Using Pythagoras’s theorem we can determine the resultant velocity, v :

v^{2} = v(h)^{2} + v(v)^{2}  latex v^{2} = (2 \times 10^{6})^{2} + (3.25 \ times 10^{7})^{2}
v^{2} = 1.05625 \times 10^{15}
v = 3.25 ^ 10^{8} ms^{-1}

You should now be able to look back at your answer and decide on whether it looks appropriate or not…

Clearly, this answer is NOT appropriate as it has turned out to be faster than the speed of light (but this is just because of a poor choice of number from the outset).

Direction of the electron upon impact (velocity has a direction after all):

The angle at which it strikes the upper plate, \theta can be determined using trigonometry:

\tan( \theta ) = \frac{2 \times 10^{6}}{3.35 \times 10^{8}}

\theta =  \tan^{-1}(\frac{2 \times 10^{6}}{3.35 \times 10^{8}})
\theta = 0.34^{\circ }

The diagram is  clearly not to scale!