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19.1Motion in an electric field

The path of an electric charge in a uniform electric field is either a straight line or a parabolic path. We have a straight line if the charge is initially at rest or if its original velocity is parallel to the electric field. In all other cases we have a parabolic path.

The diagram in Figure 19.1 shows a positively charged particle in a region of uniform electric field E. There is a potential difference V across the dotted lines X and Y.

Figure 19.1: A positively charged particle accelerates along a straight field line.

The particle will experience a force qE in the direction of the field (because the charge is positive) and so will accelerate in the direction of the field. It will move along a straight line with acceleration a=q E m. (A negative charge would move in the opposite direction.) The positive charge increases its kinetic energy as it moves to the right. The gain in kinetic energy from X to Y is equal to the work done by the electric force, and that is q V.

CHECK YOURSELF 1

A proton and an alpha particle (mass 4m p, charge +2e) are placed on the positive plate inside a parallel plate arrangement, as shown in Figure 19.2. The particles are released at the same time.

Figure 19.2: For check yourself question 1.

Which particle will reach the negative plate first?

Worked Example 19.1

A potential difference of 120 V is established between two oppositely charged parallel plates. A proton is placed on the positive plate and is then released, as shown in Figure 19.3.

Figure 19.3: For worked example 19.1.

What is the speed of the proton as it hits the negatively charged plate?

Answer

The work done by the electric field in moving the proton from one plate to the other is q V = 1.6 × 10−19 × 120 = 1.92 × 10−17 J. This goes into kinetic energy for the proton and so 1 2 m v 2=1.92×10−17 J. This gives v=2×1.92×10−17 1.67×10−27=1.5×10 5 m s−1.

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A more interesting case is when the charge enters a region of uniform electric field with an initial velocity u that is at right angles to the electric field, Figure 19.4. Consider a particle with positive charge q and mass m.

Figure 19.4: Motion in an electric field when the initial velocity is normal to the field.

When the particle enters the region of electric field it will experience a constant electric force qE directed along the field lines (i.e. downward). We are assuming that the mass is quite small so the weight of the particle is negligible compared to the electric force. Its acceleration will be a=q E m. But we have seen this situation before! This is exactly the same as a projectile that is fired horizontally (Chapter 1). Here the role of the acceleration of gravity is played by the acceleration a=q E m. So everything we learned in Chapter 1 applies here as well. Thus we know that, in Figure 19.5:

●the particle will follow a parabolic path

●the component of velocity normal to the field will remain constant: v x = u

●the component of velocity parallel to the field will increase uniformly: v y=q E m t.

Figure 19.5: The path of the particle is parabolic.

The kinetic energy of the particle increases as time goes on because the electric force is doing work on the particle. After time t, the particle will move a horizontal distance x = u t and fall a vertical distance:

y=1 2 q E m t 2

Its kinetic energy will be:

E K=1 2 m v 2=1 2 m(v x 2+v y 2)=1 2 m(u 2+(q E t m)2)

The change in kinetic energy is, therefore:

Δ E K=1 2 m(q E t m)2.

We can rewrite this as:

Δ E K=(q E)×((1 2)(q E m)t 2)=F y.

In other words, the change in kinetic energy is the work done by the electric force, as we know it should be from mechanics.

Worked Example 19.2

An electron moving at speed 6.0 × 106 m s−1 enters two oppositely charged parallel plates at a point near the lower plate, as shown in Figure 19.6. The length of the plates is 12 cm, and their separation is 4.8 cm. There is a potential difference of 60 V between the plates. The upper plate is positively charged.

Figure 19.6: For worked example 19.2.

aDetermine:

ithe magnitude and direction of the electric field in between the plates

iithe magnitude and direction of the acceleration of the electron when inside the region of electric field.

bSuggest whether the electron will hit the upper plate or not.

Answer

a iThe electric field is directed from the upper plate to the lower plate. Its magnitude is V d=60 4.8×10−2=1.25×10 3 NC−1.

iiThe acceleration is directed from the lower to the upper plate and has magnitude q E m=1.6×10−19×1.25×10 3 9.1×10−31=2.2×10 14 m s−2.

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bThe time for the electron to reach the end of the plates is 0.12 6.0×10 6=2.0×10−8 s. In this time the electron will move upwards a distance of 1 2 a t 2=1 2×2.2×10 14×(2.0×10−8)2=4.4×10−2 m.

This means the electron does not hit the upper plate.

CHECK YOURSELF 2

The potential difference between the plates in worked example 19.2 is increased. Will the electron get closer to the top plate or not?

Now consider a proton that is launched as in Figure 19.7 at an angle of 30° to the horizontal. The separation of the plates is 4.8 cm, and the potential difference is 60 kV.

Figure 19.7: A proton launched at an angle inside an electric field.

The initial vertical component of velocity is u y=6.0×10 6×sin 30°=3.0×10 6 m s−1. At the highest point the proton will cover a vertical distance found from 0=u y 2−2 a y, i.e. y=u y 2 2 a where a=e V m d is the acceleration of the proton. The acceleration is

a=1.6×10−19×60×10 3 1.67×10−27×4.8×10−2=1.20×10 14 m s−2

The vertical distance is then

y=(3.0×10 6)2 2×1.20×10 14=3.75≈3.8 cm

which means the proton will not hit the upper plate. This discussion continues in worked example 19.3.

Worked Example 19.3

The length of the plates in Figure 19.7 is 22 cm. Figure 19.8 shows the proton at its highest point.

Calculate:

athe horizontal distance covered by the proton when it is at its highest point

bthe work done on the proton by the electric field.

Figure 19.8: The proton at its highest point.

Answer

aThe time to get to the highest point is

t=u y a=3.0×10 6 1.2×10 14=2.5×10−8 s

In this time the proton moved a horizontal distance 6.0 × 106 × cos 30° × 2.5 × 10−8 ≈ 13 cm.

bThe work done on the proton by the electric field is the change in kinetic energy, which is −1 2 m u y 2

=−1 2×1.67×10−27×(3.0×10 6)2=−7.5×10−15 J.

It can also be calculated from

W=−F y=−m a y=−1.67×10−27×1.2×10 14×0.0375=−7.5×10−15 J..

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TEST YOUR UNDERSTANDING

1Why can’t a charged particle follow a circular path in a uniform electric field?

2Two parallel plates are oppositely charged. An electron is positioned on the negative plate and a proton on the positive plate (Figure 19.9).

Figure 19.9: For question 2.

The particles are released. Compare, for when the particles reach the opposite plate:

athe kinetic energy

bthe speed

cthe time to move across.

3Two parallel plates are oppositely charged. A proton (charge +e, mass m) and an alpha particle (charge +2e, mass 4m) are placed on the positively charged plate (Figure 19.10).

Figure 19.10: For question 3.

When the particles reach the opposite plate, what is the ratio of:

athe proton speed to the alpha particle speed

btime of travel of the proton to time of travel of the alpha particle?

4A proton is injected with speed 1.8 × 105 m s−1into the region between two oppositely charged parallel plates (Figure 19.11). The top plate is the negative plate, and there is a potential difference of 60 V between the plates. The plates are 8.0 cm apart and 24 cm long.

Figure 19.11: For question 4.

How far above the positive plate will the proton be when it leaves the plates?

5A proton is injected into the region between two oppositely charged parallel plates (Figure 19.12). The top plate is the positive plate, and there is a potential difference of 150 V between the plates. The plates are 4.0 cm apart. The velocity of the proton makes an angle of 30° with the horizontal.

Figure 19.12: For question 5.

aWhat is the maximum speed of the proton so that it does not hit the top plate?

bThe length of the plates is 12 cm. Show that when the proton has the maximum speed found in part a, the proton will not hit the lower plate either.

cHow would your answer to part a change if the separation of the plates was made 8.0 cm without any other changes?

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6We know that a positively charged particle launched between charged parallel plates will follow a parabolic path just like a projectile under gravity (as shown in Figure 19.13).

Figure 19.13: For question 6.

aThe mass of the charged particle and the mass of the projectile are both doubled. How do h and H change?

bThe mass of the charged particle is the original mass, but the charge is doubled. How does h change?