positive negative
a b
+ + – –
Figure 8.2.2 The electric field lines for two point charges of equal magnitude but opposite sign
Figure 8.2.4 The electric field lines between two oppositely charged parallel metal plates
Figure 8.2.3 Electric field lines around (a) two equal positive charges, and (b) two equal negative charges
positive negative
+
+ + + + + + + + + + + – – – – – – – – – – –
d E
Figure 8.2.5 Parallel plates with a uniform electric field between them
Electric field strength between parallel plates
A side-view of a uniform field between two plates is shown in Figure 8.2.5 (we have ignored edge effects).
The electric field strength E was defined as the ratio of the force F on a small positive charge q given by:
E F
= q (see in2 Physics @ Preliminary section 10.6).
It has units of force per unit charge, which in SI units is newtons per coulomb or N/C or N C–1. However, in practice neither the force between the plates nor the charge on them is easily measured. We need a more convenient expression in terms potential difference V between the plates.
The equal and opposite charges on the plates were produced by applying a potential difference V between them, using a power supply or a battery. Energy from the power supply moved the electrons from one plate to the other, resulting in equal and opposite charges on the plates. Recall that it was shown that the
Describe quantitatively the electric field due to oppositely charged parallel plates.
and is defined as:
Work = force × distance = W = F × d (see in2 Physics @ Preliminary section 4.3).
In this case, the displacement d is the separation of the parallel plates.
Rearranging this formula as follows:
F W
= d
and substituting this expression for force in the expression for electric field given previously, we obtain:
E W
= q d
×
Recall also that the work done on charges is related to the potential difference V by the following expression:
W = q × V (see in2 Physics @ Preliminary section 10.8).
Substituting this into the previous expression for electric field strength, we obtain:
E q V
= q d×
×
E V
= d
Thus, the electric field strength E is easily calculated from the plate separation d and the potential difference between them V. From this equation, you can see that an alternative unit for the electric field strength is volts per metre (V m–1).
Worked example questIon
Two parallel plates are separated by d = 1.0 cm and have a side length of L = 2.0 cm (Figure 8.2.6). A potential difference of V = 10 V is applied between them.
An electron enters halfway between the plates with a velocity of v0 = 2.7 × 106 m s–1 parallel to the plates.
Calculate:
a the electric field strength E between the plates
b the vertical force F experienced by the electron as it travels between the plates c the acceleration of the electron
d the time it takes for the electron to travel the full length of the plates e the vertical displacement of the electron just as it exits the plates
f the change in kinetic energy of the electron between its entry to and exit from the plates.
Note that the electron charge e = – 1.6 × 10–19 C, and its mass me = 9.11 × 10–31 kg.
solutIon
a The electric field strength E is given by:
E V
= =d 10 = − −
0 010 1000 1 1
. V m (or N C )
Solve problem and analyse information using:
F = qvBsinθ F = qE and
E V
= d
+ + + + + + + + + + + – – – – – – – – – – –
e– d
E
L
Figure 8.2.6 Two parallel plates with a uniform electric field
b The vertical force is given by:
F = e × E = 1.6 × 10–19 × 1000 = 1.6 × 10–16 N upwards
c The acceleration can be obtained from Newton’s second law (see section 3.4 of in2 Physics @ Preliminary):
a F
=m = ×
× −− = 1.76 × 1014 m s–2 = 1.8 × 1014 m s–2 downwards
e
1 6 10 9 11 10
16 31
. .
d The motion of the electron is similar to projectile motion (see Module 1 ‘Space’).
There is only acceleration in the vertical direction. We are concerned with the horizontal component of the velocity, which remains constant. Therefore, the time t to travel the length L of the plates is given by:
t L
= =v
× = × −
0 6
0 020 9
2 7 10. 7.4 10 s .
e The electron only accelerates vertically, so we use the acceleration from part c and one of the SUVAT equations (see in2 Physics @ Preliminary section 1.3) to work out the vertical displacement s given by:
s ut= +1at 2
2
where u is the initial vertical velocity = 0 m s–1, t is the time to travel the length of the plates (from part d), and a is the vertical acceleration = 1.8 × 1014 m s–2 (from part c). Therefore, the displacement is given by:
s= + × ×0 1 × × − = =
2 1 8 10. 14 ( .7 4 10 9 2) 0 0049. m 4 9. mm
f The change in kinetic energy is due to the vertical deflection of the electron by the electric field. This is work done by the field on the electron, and is given by:
W e= × ∆V
where ΔV is the change in potential that the electron experiences due to its vertical deflection s. The relationship between potential difference and the displacement s is given by:
∆V E s= ×
Combining this with the above expression for work gives:
W = e × E × s = 1.6 × 10–19 × 1000 × 0.0049 = 7.8 × 10–19 J
CheCkpoInt 8.2
1 Explain the meaning of the arrows and spacing of lines when drawing electric field lines.
2 Calculate the electric field strength at the location of a charge of 1.28 × 10–18 C that experiences a force of 1.1 × 10–18 N.
3 Sketch the electric field lines around:
a a point positive charge
b two oppositely charged parallel metal plates.
8.3 Charges moving in a magnetic field
A charged particle moving in a magnetic field will experience a force, as discussed in Module 2 ‘Motors and generators’. Consider a magnetic field directed into the pages (see Figure 8.3.1) and a positively charged particle that enters the field from the left. The particle will experience a force as given by the right-hand palm rule (refer to Module 2 ‘Motors and generators’). The force F on the particle changes direction every time the particle changes direction, such that the resulting motion is circular. This circular motion is known as cyclotron motion.
A negatively charged particle will undergo circular motion in the opposite sense.
The magnitude of the force F on a charge q that moves with a speed v at right angles to a magnetic field B is given by:
F = qvB
The speed of the charged object remains constant even though its direction changes. This means that the magnetic force simply changes the direction of the particle without adding or subtracting energy, provided that the magnetic field is constant. In general, the magnitude of the force on a charged particle with a velocity at an angle θ with respect to the magnetic field is given by:
F = qvBsinθ
Note that a particle that is parallel to the field (θ = 0°) does not experience a force. A charged particle that enters a magnetic field at an angle other than θ = 0° or 90° travels in a spiral along the magnetic field (also known as helical motion), as shown in Figure 8.3.2. However, the component of the velocity at right angles to the field is still circular.
A particle of mass m with charge q enters a uniform magnetic field B at an angle θ and a speed v. Recall from the Module 1 ‘Space’ that the force directed towards the centre of circular motion is known as the centripetal force. In this case, the centripetal force is the magnetic force; that is:
centripetal force = magnetic force mv
r2 =qvBsinθ where r is the radius of curvature of the particle.
Rearranging this expression, we obtain the following expression for the radius:
r mv
=qB sinθ
Identify that moving charged particles in a magnetic field experience a force.
Describe quantitatively the force acting on a charge moving through a magnetic field:
F = qvBsinθ
Solve problem and analyse information using:
F = qvBsinθ F = qE and
E V
= d + F
+q v
r
Figure 8.3.1 A positively charged particle enters into a magnetic field directed into the page and undergoes anticlockwise circular motion.
+ v
B
θ v
B
a b
F
+
v sinθ
v cosθ B
c
+
Figure 8.3.2 (a) A proton enters a magnetic field at an angle θ. (b) The component of the velocity at right angles to the field undergoes circular motion.
(c) The total motion is helical along the magnetic field.
Worked example questIon
Protons are occasionally ejected from the Sun at high speeds towards the Earth. These can be caught in the Earth’s magnetic field and are trapped along magnetic field lines in a region known as the Van Allen belt. A proton travelling at a speed of v = 1.0 × 107 m s–1 strikes the Earth’s magnetic field at an angle of 30° at a distance of 3000 km above the surface where the magnetic field B = 3.5 × 10–5 T. Assume that the Earth’s magnetic field at this distance is uniform. The charge on a proton is 1.6 × 10–19 C and it has a mass of 1.67 × 10–27 kg.
a Calculate the magnitude of the force on the proton.
b Calculate the radius of curvature of the proton.
c Determine if the cyclotron motion of the proton will cause it to come in contact with the Earth.
solutIon
a The magnitude of the force is given by:
F qvB= sinθ 1 6 10= ×. −19× ×1 0 10 3 5 10. 7× ×. −5sin30°=2.8 × 10 N–17 b The radius of curvature r is given by:
r mv
=qB = × × ×
× × ×
−
− −
sin
. .
. .
θ
1 67 10 1 0 10 1 6 10 3 5 10
27 7
19 5ssin .
30 6 0 103
° = × m
c No, the proton’s cyclotron motion will not cause it to come in contact with the Earth’s surface, since the cyclotron radius is 6 km and the proton is 3000 km above the surface.
CheCkpoInt 8.3
1 Outline what will happen to a negative charge moving to the left at right angles to a magnetic field that is directed out of the page.
2 Calculate the magnitude of the force experienced by an electron that travels at right angles to a magnetic field of 2 T at a speed of 3 m s–1.
8.4 thomson’s experiment
In 1897 Joseph John (JJ) Thomson not only provided a definitive resolution to the debate about whether cathode rays were particles or electromagnetic waves, but he also measured the charge to mass ratio of the main constituent of cathode rays—the electron.
Thomson’s cathode ray tube is shown schematically in Figure 8.4.1. Cathodes rays (electrons) were accelerated from the cathode to the anode, which consisted of two anodes aligned along the axis of the tube and separated by a small distance.
Each anode had a horizontal slit cut into it so that cathode rays could pass through.
The separation of the anodes produced a flat beam that passed between two parallel plates and struck the end of the tube, which had a fluorescent screen painted on the inside surface. The beam produced a well-defined narrow horizontal line on the screen.
Outline Thomson’s experiment to measure the charge/mass ratio of an electron.
The parallel plates deflected the beam vertically upwards by placing a positive charge on the top plate. A magnetic field (using an electromagnet outside the tube) was applied at right angles to the electric field and the direction of the beam. The direction of the magnetic field was such that it deflected the beam downwards.
Adjusting the magnetic and electric forces such that they were equal and opposite resulted in the beam passing through undeflected.
We now quantitatively describe this experiment and how it leads to the charge to mass ratio of the electron.
Equating the electric to the magnetic force on a particle with a charge q and speed v in a magnetic field B and an electric field E, we obtain the following expression: Magnetic force = electric force
qvB = qE From this we obtain an expression for the speed:
v E
= B
Recall that the relationship between the magnetic force and centripetal force (see section 8.3) is given by: mv
r2 =qvB
This enables an expression for charge to mass ratio to be obtained:
q m
v
=rB
Substituting the expression for v into this equation:
q m
E
=rB2
Thomson was able to calculate the radius of curvature r from the deflection of the beam on the fluorescent screen when the electric field was switched off and magnetic field switched on. He could calculate the magnitude of the electric field E because he knew the spacing d of the plates and the potential difference V between them, and made use of the relationship E=V/d. Finally, knowing the number of turns in the wires of the electromagnet and the current flowing through it, he was able to calculate the magnitude of the magnetic field B.
Thomson found that the charge to mass ratio always came to:
q
m=1 76 10. × 11 C kg−1
regardless of the cathode material, indicating that a fundamental particle was being emitted. This, in essence, marks the discovery of the electron. In 1891, the Irish physicist George Johnstone Stoney (1826–1911) suggested that the fundamental unit of electricity be called an electron—6 years before Thomson’s publication of his now famous experiment.
electromagnet charged plates
fluorescent screen voltage
cathode anodes
Figure 8.4.1 Thomson’s cathode ray tube apparatus for measuring the charge to mass ratio of electrons
CheCkpoInt 8.4
1 Explain the purpose of Thomson’s experiment.
2 Describe how Thomson’s experiment obtained the charge to mass ratio of the electron.
8.5 applications of cathode rays
A cathode ray oscilloscope (or simply oscilloscope or CRO) is a device used to measure the variation of voltage in time across an electrical component. Both the CRO and TV have elements in common with Thomson’s original cathode ray tube apparatus. Here we are not referring to the new plasma or LCD television sets, but the older style scanning electron beam sets.