A charged particle +q enters an electric field with a velocity v at right angles to the direction of the field (see Figure 17.12). The particle experiences a force at right angles to its initial direction. If the field direction is vertical and the initial direction is horizontal, the particle will follow a parabolic path as it passes through the field.
path of particle, charge q mass m velocity v
region of uniform electric field, strength E
Figure 17.12
The analysis of the motion is similar to that of a particle of mass m moving in a gravitational field with a uniform velocity in one direction and a uniform acceleration in a perpendicular direction (as described for projectile motion in Topic 3).
Example
A particle of mass 6.7 × 10−27 kg and charge +3.2 × 10−19 C is placed near the top plate in an electric field similar to that shown in Figure 17.11. The potential difference across the plates is 600 V and the separation of the plates is 12 mm. Determine
(a) the force on the particle, (b) the acceleration of the particle,
(c) the speed of the particle when it reaches the bottom plate.
(a) force = Eq = Vq/d = (600 × 3.2 × 10−19)/12 × 10−3 = 1.6 × 10−14 N (b) acceleration = force/mass = 2.3 × 10−12/6.7 × 10−27 = 2.4 × 1012 m s−2 (c) v2 = 0 + 2as = 2 × 2.4 × 1012 × 12 × 10−3 so speed v = 2.4 × 105 m s−1
Now it’s your turn
6 An electron enters an electric field similar to that shown in Figure 17.12 with a horizontal velocity of 7.0 × 107 m s−1. The horizontal length of the plates is 2.0 cm. The electric field strength is 3.2 × 105 V m−1. Calculate the vertical displacement of the electron for the time it is between the plates.
Summary
● Insulators may be charged by friction.
● Like charges repel; unlike charges attract each other.
● When charged objects are placed near conductors, they cause a redistribution of charge in the conductor, thereby inducing charges.
● An electric field is a region of space where a stationary charge experiences a force.
● Electric field strength is the force per unit positive charge: E = F/Q
● The electric field between parallel plates is uniform. The field strength is given by:
E = V/d
● The motion of a charged particle moving in the direction of the electric field can be analysed using the equations of constant acceleration.
● The motion of a charged particle moving with a uniform velocity in one direction and a uniform acceleration in a perpendicular direction is parabolic.
Examination style questions
1 Calculate the speed of an electron accelerated from rest through a distance of 40 mm by a uniform electric field of 3.0 × 103 N C−1.
2 An electron starts from rest from the bottom plate in Fig. 17.11. The potential difference across the plates is 1600 V and the separation of the plates is 15 mm. Calculate the time taken for the electron to reach the top plate.
3 For the electron in question 2 calculate:
a the work done by the field on the electron, b the gain in kinetic energy,
c the speed of the electron.
4 a Two horizontal metal plates are connected to a power supply, as shown in Fig. 17.13.
1.2 kV+ then closed so that a potential difference of 1.2 kV is applied across the plates.
i On a copy of Fig. 17.13, draw six field lines to represent the electric field between the metal plates. [2]
ii Calculate the electric field strength E between the
plates. [2]
b The switch S is opened and the plates lose their charge.
Two very small metal spheres A and B joined by an insulating rod are placed between the metal plates as shown in Fig. 17.14.
17 Electric fi elds
Sphere A has charge –e and sphere B has charge +e, where e is the charge of a proton. The length AB is 15 mm. The rod is supported at its centre C so that the rod is horizontal and in equilibrium.
The switch S is then closed so that the potential difference of 1.2 kV is applied across the plates.
i There is a force acting on A due to the electric field between the plates. Show that this force is
4.8 × 10–15 N. [2]
ii The insulating rod joining A and B is fixed in the position shown in Fig. 17.14. Calculate the torque of
the couple acting on the rod. [3]
iii The insulating rod is now released so that it is free to rotate about C. State and explain the position of the
rod when it comes to rest. [2]
Cambridge International AS and A level Physics, 9702/23 Oct/Nov 2013 Q 7 5 Two vertical parallel metal plates are situated 2.50 cm
apart in a vacuum. The potential difference between the plates is 350 V, as shown in Fig. 17.15.
350 V
2.50 cm
electron + −
Fig. 17.15
An electron is initially at rest close to the negative plate and in the uniform electric field between the plates.
a i Calculate the magnitude of the electric field
between the plates. [2]
ii Show that the force on the electron due to the electric field is 2.24 x 10–15 N. [2]
b The electron accelerates horizontally across the space between the plates. Determine
i the horizontal acceleration of the electron , [2]
ii the time to travel the horizontal distance of 2.50 cm
between the plates. [2]
c Explain why gravitational effects on the electron need not be taken into consideration in your
calculation in b. [2]
Cambridge International AS and A level Physics, 9702/21 May/June 2009 Q 6
6 a Define electric field strength. [1]
b Two flat parallel metal plates, each of length 12.0 cm, are separated by a distance of 1.5 cm, as shown in Fig. 17.16.
The space between the plates is a vacuum. The potential difference between the plates is 210 V. The electric field may be assumed to be uniform in the region between the plates and zero outside this region.
Calculate the magnitude of the electric field strength
between the plates. [1]
c An electron initially travels parallel to the plates along a line mid-way between the plates, as shown in
Fig. 17.16. The speed of the electron is 5.0 × 107 m s–1. For the electron between the plates,
i determine the magnitude and direction of its
acceleration, [4]
ii calculate the time for the electron to travel a horizontal distance equal to the length of
the plates. [1]
d Use your answers in c to determine whether the electron will hit one of the plates or emerge from
between the plates. [3]
Cambridge International AS and A level Physics, 9702/02 May/June 2007 Q 2
19 Current of electricity
AS Level
Starting points
● This topic considers fundamental ideas about electric charge and electric current.
● Examples of electric currents are in household wiring and electrical appliances.
● A potential difference is required for energy changes to occur in a circuit.
● Resistance controls the flow of charge in a circuit.
19.1 Charge and current
All matter is made up of tiny particles called atoms, each consisting of a positively charged nucleus with negatively charged electrons moving around it.
Charge is measured in units called coulombs (symbol C). The charge on an electron is −1.6 × 10−19 C. Normally atoms have equal numbers of positive and negative charges, so that their overall charge is zero. But for some atoms it is relatively easy to remove an electron, leaving an atom with an unbalanced number of positive charges. This is called a positive ion.
Robert Millikan in 1912 performed an experiment to determine the charge on an electron using charged oil droplets (see Topic 17). The experimental result showed that, no matter what the charge on the droplets, it seemed to occur only in integral multiples of e, the charge on an electron. The conclusion was that charge is not continuous but quantised, or exists only in discrete amounts. The photon is another example of a quantised physical quantity, introduced in Topic 25.
Atoms in metals have one or more electrons which are not held tightly to the nucleus. These free (or mobile) electrons wander at random throughout the metal.
However, when a battery is connected across the ends of the metal, the free electrons drift towards the positive terminal of the battery, producing an electric current.
The size of the electric current is given by the rate of flow of charge. Electric current is an SI base quantity (see Topic 1). The SI base unit of current is the ampere (or amp for short), with symbol A. The SI units of all the other electrical quantities are derived from the SI base units.
By the end of this topic, you will be able to:
19.1 (a) understand that electric current is a flow of charged carriers
(b) understand that the charge on charge carriers is quantised
(c) define the coulomb (d) recall and use Q = It
(e) derive and use, for a current-carrying conductor, the expression I = Anvq, where n is the number density of charge carriers
19.2 (a) define potential difference and the volt (b) recall and use V = W/Q
(c) recall and use P = VI, P = I2R 19.3 (a) define resistance and the ohm
(b) recall and use V = IR
(c) sketch and discuss the I–V characteristics of a metallic conductor at constant temperature, a semiconductor diode and a filament lamp (d) state Ohm’s law
(e) recall and use R = ρl/A
19 Current of electricity
nucleus
electrons
Figure 19.1 Atoms consist of a positively charged nucleus with negative electrons outside.
neutral atom
positive ion
electron
Figure 19.2 An atom with one or more electrons missing is a positive ion.
A current of 3 amperes means that 3 coulombs pass a point in the circuit every second.
In 5 seconds, a total charge of 15 coulombs will have passed the point. So, charge = current × time
or Q = It
where the charge Q is in coulombs when the current I is in amperes and the time t is in seconds. This gives a definition of the coulomb as follows.
The coulomb is that charge passing a point in a circuit when there is a current of one ampere for one second.
Example
The current in the filament of a torch bulb is 0.03 A. How much charge flows through the bulb in 1 minute?
Using Q = It, Q = 0.03 × 60 (remember the time must be in seconds), so Q = 1.8 C.
Now it’s your turn
1 Calculate the current when a charge of 240 C passes a point in a circuit in a time of 2 minutes.
2 In a silver-plating experiment, 9.65 × 104 C of charge is needed to deposit a certain mass of silver. Calculate the time taken to deposit this mass of silver when the current is 0.20 A.
3 The current in a wire is 200 mA. Calculate:
(a) the charge which passes a point in the wire in 5 minutes, (b) the number of electrons needed to carry this charge
(electron charge e = −1.6 × 10−19 C).