CAIE A-Level Physics 18 Electric Fields

Practise analysing electric fields, force and potential, including field lines, uniform plates, Coulomb’s law, gradients and point-charge energy.

Exam points

represent electric fields and field lines and calculate force or field strength
analyse uniform fields and charged-particle motion between plates
apply Coulomb’s law and point-charge field relationships
calculate electric potential, potential gradient and point-charge potential energy

Covered syllabus points

18.1 Electric fields and field lines 18.2 Uniform electric fields 18.3 Electric force between point charges 18.4 Electric field of a point charge 18.5 Electric potential

[Maximum number: 8]

(a)

Define electric field.

[ 2 ]

(b)

Fig. 5.1 shows two parallel conducting plates that are in a vacuum. The plates are separated by a distance of 6.7 cm and have a potential difference (p.d.) of 430 V between them.

Fig. 5.1

[ 6 ]

(i)

On Fig. 5.1, draw four field lines to represent the electric field between the plates.

[ 2 ]

(ii)

Determine the strength E of the electric field between the plates.

E=\ldots \ldots \ldots \ldots \ldots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ~ \mathrm{NC}^{-1}

[ 2 ]

E=V / d

C1

E=430 / 0.067
$=6.4 \times 10^{3} \mathrm{~N} \mathrm{C}^{-1}$

A1

(iii)

An electron travels at a speed of $2.6 \times 10^{7} \mathrm{~ms}^{-1}$ towards the region between the plates, as shown in Fig. 5.1.

On Fig. 5.1, draw the path of the electron as it moves between and beyond the plates.

[ 2 ]

[Maximum number: 5]

Two parallel metal plates X and Y are separated by a distance of 0.041 m , as shown in Fig. 6.1.

Fig. 6.1

There is a vacuum between the plates. An electron is at rest at the centre of plate X.
A potential difference (p.d.) of 58 kV is applied across the plates. This causes the electron to accelerate towards plate Y.

(a)

On Fig. 6.1, use the symbols + and - to indicate which of plates X and Y is the positive plate and which is the negative plate.

[ 1 ]

(b)

Calculate the electric field strength E between the plates. Give a unit with your answer.
E= unit

[ 2 ]

E=V / x

C1

$\begin{aligned} =\left(58 \times 10^{3}\right) / 0.041 =1.4 \times 10^{6} \mathrm{NC}^{-1} \end{aligned}$

A1

(c)

Determine the acceleration of the electron.
acceleration = $\mathrm{ms}^{-2}$ [2]

[ 2 ]

m a=e E

C1

$\begin{aligned} a =\left(1.60 \times 10^{-19} \times 1.41 \times 10^{6}\right) /\left(9.11 \times 10^{-31}\right) =2.5 \times 10^{17} \mathrm{~m} \mathrm{~s}^{-2} \end{aligned}$

A1

[Maximum number: 3]

An isolated metal sphere of radius r is charged so that the electric field strength at its surface is $E_{0}$ .

On Fig. 6.1, sketch the variation of the electric field strength E with distance x from the centre of the sphere. Your sketch should extend from x=0 to x=3 r.

Fig. 6.1

from x=0 to x=r: E=0

B1

from x=r to x=3 r : curve with negative gradient of decreasing magnitude passing through ( $r, E_{0}$ )

B1

line passing through ( $2 r, E_{0} / 4$ ) and ( $3 r, E_{0} / 9$ )

B1

[Maximum number: 7]

(a)

Define electric field.

[ 1 ]

(b)

An isolated uniform conducting sphere has mass M and charge Q.

The gravitational field strength at the surface of the sphere is g.
The electric field strength at the surface of the sphere is E.

[ 3 ]

(i)

Show that

\frac{M}{Q}=\alpha \frac{g}{E}

where $\alpha$ is a constant.

[ 3 ]

$g=G M / r^{2}$

M1

$E=Q / 4 \pi \varepsilon_{0} r^{2}$

M1

algebra showing the elimination of r leading to $M / Q=\left(1 / 4 \pi G \varepsilon_{0}\right)(g / E)$

A1

(c)

Assume that the Earth is a uniform conducting sphere of mass $5.98 \times 10^{24} \mathrm{~kg}$ . The surface of the Earth carries a charge of $-4.80 \times 10^{5} \mathrm{C}$ that is evenly distributed.

[ 3 ]

(i)

Use the information in (b) to determine the electric field strength at the surface of the Earth. Give a unit with your answer.
electric field strength = unit

[ 2 ]

$\begin{aligned} E =\alpha g Q / M =\left(1.35 \times 10^{20} \times 9.81 \times 4.80 \times 10^{5}\right) /\left(5.98 \times 10^{24}\right) \end{aligned}$

C1

$=106 \mathrm{~N} \mathrm{C}^{-1}$ or $106 \mathrm{~V} \mathrm{~m}^{-1}$

A1

(ii)

State how the direction of the electric field at the surface of the Earth compares with the direction of the gravitational field.

[ 1 ]

All question bank results loaded

Covered syllabus points

18. Electric fields question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

18. Electric fields question 2

Question (a)

Question (b)

Question (c)

18. Electric fields question 3

18. Electric fields question 4

Question (a)

Question (b)

Question (i)

Question (c)

Question (i)

Question (ii)