Electrostatics — Physical Sciences | Learn on Stanmorephysics

4 exam-style question sets on Electrostatics (Paper 1), each with a hint and a fully worked answer. The app holds 20 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Electric field & Coulomb's law on a lineCoulomb's law & equilibriumField of a sphere; net force and friction on a benchCoulomb's law: force balance on a line

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Question 1

Electric field & Coulomb's law on a lineComprehension14 marksPaper 1

Small sphere A, carrying a charge of +4 nC, is held in a fixed position on an insulated bench. Point P lies 5 cm to the left of A.

Sphere B, carrying a charge of −4 nC, is then fixed 6 cm to the right of A. Finally, a third small sphere C is fixed 3 cm to the right of B so that A, B and C lie on the same straight line, as shown in the diagram.

Sphere C experiences a net electrostatic force of magnitude 3,2 × 10⁻⁴ N directed to the LEFT.

aCalculate the magnitude of the electric field at point P due to sphere A ONLY. (3)
bDraw the resultant electric field pattern due to spheres A and B only. (3)
cState Coulomb's law in words. (2)
dIs the charge on sphere C POSITIVE or NEGATIVE? Explain the answer by referring to the directions of the forces acting on C. (1)
eCalculate the magnitude of the charge on sphere C. (5)

This question has a diagram, shown in the app.

Hint

Use E = kQ/r² for the field; for the last part write the net force on C as the difference between the two Coulomb forces and solve for the unknown charge.

Worked answer

a) $E = \frac{k Q}{r ^{2}} = \frac{( 9 , 0 \times 1 0 ^{9} ) ( 4 \times 1 0 ^{- 9} )}{( 0 , 05 ) ^{2}}$ ✓✓
$E = 1, 44 \times 1 0^{4}$ N·C⁻¹ ✓ (formula ✓ substitution ✓ answer+unit ✓)

b) Full-mark drawing must show: field lines leaving the positive sphere A radially and entering the negative sphere B ✓; lines curving from A to B in the region between and around the spheres ✓; correct arrow directions (away from A, towards B), lines never crossing and never touching the spheres' interiors ✓.

c) The magnitude of the electrostatic force exerted by one point charge on another is directly proportional to the product of (the magnitudes of) the charges and inversely proportional to the square of the distance between them. ✓✓

d) POSITIVE. ✓ (B, the nearer sphere, must attract C to the left while A repels it to the right; only a positive charge on C gives a net force to the left.)

e) C is attracted towards B (left) and repelled by A (right). Taking left as positive:
$F_{n e t} = \frac{k Q _{B} Q _{C}}{r _{B C}^{2}} - \frac{k Q _{A} Q _{C}}{r _{A C}^{2}}$ ✓
$3, 2 \times 1 0^{- 4} = (9, 0 \times 1 0^{9}) (4 \times 1 0^{- 9}) Q_{C} (\frac{1}{( 0 , 03 ) ^{2}} - \frac{1}{( 0 , 09 ) ^{2}})$ ✓✓
$3, 2 \times 1 0^{- 4} = 3, 56 \times 1 0^{4} Q_{C}$ ✓
$Q_{C} = 9, 0 \times 1 0^{- 9}$ C $= 9, 0$ nC ✓

Modelled on Nov 2024 Paper 1, Question 7 — same skills, new scenario.

Question 2

Coulomb's law & equilibriumApplication2 marksPaper 1

A small sphere of mass m carrying charge +q floats stationary a distance d directly above an identical fixed sphere that also carries charge +q. Only the gravitational force and the electrostatic force act on the upper sphere.

Which ONE of the following expressions gives the charge q?

Hint

Set the Coulomb force equal to the weight and solve for q.

Worked answer

C. In equilibrium the Coulomb repulsion balances the weight: $\frac{k q ^{2}}{d ^{2}} = m g$ , so $q^{2} = \frac{m g d ^{2}}{k}$ and $q = d \frac{m g}{k}$ . A puts $k$ in the numerator and misplaces $d$ . B divides by $d^{2}$ instead of multiplying. D is $q^{2}$ , not $q$ — the square root was omitted.

Modelled on Nov 2024 Paper 1, Question 1.7 — same skill, new scenario.

Question 3

Field of a sphere; net force and friction on a benchEvaluation15 marksPaper 1

A small sphere X, carrying a charge of +2 × 10⁻⁷ C, rests at a fixed position on a horizontal wooden bench. Point T lies a distance $d$ to the RIGHT of X on the bench. The magnitude of the electric field at point T due to sphere X is 7,2 × 10⁵ N·C⁻¹.

A sphere Y, carrying a charge of −6 × 10⁻⁷ C, is now fixed at point T, and a sphere Z, carrying a charge of +4 × 10⁻⁷ C, is fixed 0,08 m to the LEFT of sphere X, so that Z, X and Y lie on the same straight line.

aDescribe the term electric field. (2)
bDraw the electric field pattern due to sphere X alone. (2)
cShow, by means of a calculation, that d = 0,05 m. (3)
dWhen sphere X is released, the net (resultant) force acting on it is measured as 0,42 N. Is the bench frictionless? Choose from YES or NO. Explain the answer by means of a calculation. (6)
eSphere Y is brought into contact with sphere X and then placed back at point T. How will the magnitude of the force that sphere X now exerts on sphere Y be affected? Choose from INCREASES, DECREASES or REMAINS THE SAME. Explain the answer. (2)

This question has a diagram, shown in the app.

Hint

First find each electrostatic force on X separately and decide its direction, then compare their resultant with the measured net force.

Worked answer

a) An electric field is a region of space in which an electric charge experiences an electrostatic force. ✓✓

b) Field lines radiate radially OUTWARD from sphere X (positive charge) ✓, evenly spaced, starting on the sphere's surface and never crossing ✓.

c) $E = \frac{k Q}{d ^{2}}$ ✓
$7, 2 \times 1 0^{5} = \frac{( 9 , 0 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 7} )}{d ^{2}}$ ✓
$d^{2} = 2, 5 \times 1 0^{- 3} \Rightarrow d = 0, 05$ m ✓

d) Force of Y on X: $F = \frac{k Q _{X} Q _{Y}}{r ^{2}} = \frac{( 9 , 0 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 7} ) ( 6 \times 1 0 ^{- 7} )}{( 0 , 05 ) ^{2}} = 0, 43$ N ✓ (attraction — to the right, towards Y) ✓
Force of Z on X: $F = \frac{( 9 , 0 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 7} ) ( 4 \times 1 0 ^{- 7} )}{( 0 , 08 ) ^{2}} = 0, 11$ N ✓ (repulsion — to the right, away from Z) ✓
Net electrostatic force $= 0, 432 + 0, 1125 = 0, 54$ N to the right ✓
The measured net force (0,42 N) is SMALLER than 0,54 N, so a friction force of about 0,12 N must also act on X. NO, the bench is not frictionless. ✓

e) DECREASES. ✓ After contact the identical spheres share the total charge: $\frac{( + 2 \times 1 0 ^{- 7} ) + ( - 6 \times 1 0 ^{- 7} )}{2} = - 2 \times 1 0^{- 7}$ C each — both charge magnitudes are now smaller (2 instead of 2 × 6 product scale), so by Coulomb's law the force is smaller. ✓

Modelled on Nov 2025 Paper 1, Question 7 — same skills, new scenario.

Question 4

Coulomb's law: force balance on a lineApplication2 marksPaper 1

Three point charges A, B and C are fixed on a straight line, with B between A and C. The distance AB = d and the distance BC = d. The magnitude of the charge on B is q.

The net electrostatic force on charge A is ZERO. The magnitude of the charge on C is ...

This question has a diagram, shown in the app.

Hint

C is twice as far from A as B is — apply the inverse-square law.

Worked answer

D. For zero net force on A: $\frac{k q Q _{A}}{d ^{2}} = \frac{k Q _{C} Q _{A}}{( 2 d ) ^{2}}$ , so $Q_{C} = 4 q$ — the charge twice as far away must be $2^{2} = 4$ times bigger. A inverts this ratio. B and C use the distance factor 2 without squaring it, forgetting that Coulomb's law is an inverse-SQUARE law. Modelled on Nov 2025 Paper 1, Question 1.7 — same skills, new scenario.

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Grade 12 Physical Sciences: Electrostatics

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