CAIE A-Level Physics A2 18.3 Electric Force Between Point Charges Questions

Practise applying Coulomb’s law to calculate and interpret force between point charges, rearranging it and reasoning about sign, direction and inverse-square dependence.

Exam points

apply spherical-conductor and Coulomb-law models to calculate and interpret electric force between point charges
rearrange Coulomb relationships and reason about sign, direction and inverse-square dependence

Covered syllabus points

18.3.1—That, for a point outside a spherical conductor, the charge on the sphere 18.3.2—Coulomb’s law F = Q1Q2 / (4πε0 r 2) for the force between two point charges

[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
from x=r to x=3 r : curve with negative gradient of decreasing magnitude passing through ( $r, E_{0}$ )
line passing through ( $2 r, E_{0} / 4$ ) and ( $3 r, E_{0} / 9$ )

[Maximum number: 4]

(a)

State Coulomb's law.

[ 2 ]

(b)

Two identical oil droplets are in a vacuum. The centres of the droplets are a distance of $3.8 \times 10^{-6} \mathrm{~m}$ apart. The droplets have equal charge and exert an electric force on each other of magnitude $6.3 \times 10^{-17} \mathrm{~N}$ .

Determine the magnitude of the charge on each droplet.
charge = C

[ 2 ]

$F=Q^{2} / 4 \pi \varepsilon_{0} X^{2}$ $6.3 \times 10^{-17}=Q^{2} /\left[4 \pi \times 8.85 \times 10^{-12} \times\left(3.8 \times 10^{-6}\right)^{2}\right]$
charge $=3.2 \times 10^{-19} \mathrm{C}$

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)