CAIE A-Level Physics 18.5 Electric Potential

Practise calculating electric potential and potential energy for point charges, using potential gradients for field strength and relating signs to work and energy.

Exam points

define and calculate electric potential and point-charge potential, including sign and reference
use potential gradient to determine electric field strength and direction
calculate electric potential energy of two point charges and relate it to work and energy changes

Covered syllabus points

18.5.1—Electric potential at a point as the work done per unit positive charge in 18.5.2—The fact that the electric field at a point is equal to the negative of 18.5.3—V = Q / (4πε0 r) for the electric potential in the field due to a point 18.5.4—The concept of electric potential leads to the electric potential energy of

[Maximum number: 6]

(a)

Define what is meant by electric potential at a point.

[ 2 ]

(b)

In an $\alpha$ -particle scattering experiment, $\alpha$ -particles are directed towards a thin film of gold, as illustrated in Fig. 9.1.

Fig. 9.1

gold film

The apparatus is in a vacuum.
The gold-197 ( ${ }_{79}^{197} \mathrm{Au}$ ) nuclei in the film may be considered to be fixed point charges.
The $\alpha$ -particles emitted from the source each have an energy of 4.8 MeV .
Calculate:

[ 4 ]

(i)

the distance d of closest approach of an $\alpha$ -particle to a gold nucleus.

\begin{aligned} & d= \\ & \text { m [4] } \end{aligned}

[ 4 ]

$E_{\mathrm{P}}=\mathrm{Qq} / 4 \pi \varepsilon_{0} d$

C1

Q=79 e and q=2 e

C1

$7.68 \times 10^{-13}=\left(79 \times 2 \times\left\{1.60 \times 10^{-19}\right\}^{2} /\left(4 \pi \times 8.85 \times 10^{-12} \times d\right)\right.$

C1

$d=4.7 \times 10^{-14} \mathrm{~m}$

A1

[Maximum number: 2]

The electric potential gradient is related to the electric field.

Use data from Fig. 4.2 to state the value of x at which the magnitude of the electric potential gradient is maximum. Give a reason for the value you have chosen.

field strength or E is potential gradient or field strength is rate of change of (electric) potential

M1

(field strength) maximum at $x=6 \mathrm{~cm}$

A1

[Maximum number: 5]

Two small solid metal spheres A and B have equal radii and are in a vacuum. Their centres are 15 cm apart.
Sphere A has charge +3.0 pC and sphere B has charge +12 pC . The arrangement is illustrated in Fig. 5.1.

Fig. 5.1

Point P lies on the line joining the centres of the spheres and is a distance of 5.0 cm from the centre of sphere A .

(a)

Calculate the electric potential at point P.

[ 2 ]

potential $=8.99 \times 10^{9}\left\{\left(3.0 \times 10^{-12}\right) /\left(5.0 \times 10^{-2}\right)+\left(12 \times 10^{-12}\right) /\left(10 \times 10^{-2}\right)\right\}$

\text { = } 1.62 \mathrm{~V}

(b)

A silver-107 nucleus $\left({ }_{47}^{107} \mathrm{Ag}\right)$ has speed v when it is a long distance from point P .

Use your answer in (c) to calculate the minimum value of speed v such that the nucleus can reach point P.
speed = $\mathrm{ms}^{-1}$

[ 3 ]

$\frac{1}{2} m v^{2}=q V$ $E_{\mathrm{K}}=1 / 2 \times 107 \times 1.66 \times 10^{-27} \times v^{2}$ $q V=47 \times 1.60 \times 10^{-19} \times 1.62$ $v^{2}=1.37 \times 10^{8}$ $v=1.2 \times 10^{4} \mathrm{~m} \mathrm{~s}^{-1}$

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Covered syllabus points

18.5 Electric potential question 1

Question (a)

Question (b)

Question (i)

18.5 Electric potential question 2

18.5 Electric potential question 3

Question (a)

Question (b)