Grade 11 Physical Sciences · Electricity and Magnetism

Find net electrostatic force or field

Add the forces or fields from two or more charges at one point to find the net force or field and its direction.

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Grade 11 and 12 Electrostatics Exam QuestionMiss Martins Maths and Science

Coulomb's Law (5 of 7) Force from Three Charges in a Straight LineStep by Step Science

Coulomb's Law (7 of 7) Force on Three Charges Arranged in a Right TriangleStep by Step Science

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Question 1.7 · November 2025 · Paper 1 · National (DBE) · 2 marks

Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g. 1.11 E.

1.7 Three point charges,

Q_{1}

Q_{2}

and

Q_{3}

, are fixed in a straight line.

Q_{1}

is r metres from

Q_{2}

, while

Q_{3}

is 2r metres from

Q_{2}

, as shown in the diagram below.

The magnitude of the charge on

Q_{2}

is q. The net electrostatic force on charge

Q_{1}

is zero.

What is the magnitude of charge

Q_{3}

in terms of q?(2) Diagram for question 1.7

A $\frac{1}{9}$ q
B $\frac{1}{3}$ q
C 3q
D 9q

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: find net electrostatic force or field

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3. Question 7.3.3 · November 2024 · Paper 1 · National (DBE) · 5 marks
P is a $+ 2$ nC point charge. X is a point $6$ cm away from charge P, as shown in the diagram below.
7.3 A third point charge T is placed $2$ cm to the right of S, as shown in the diagram below. Point charge T experiences a net electrostatic force of $2, 5 \times 1 0^{- 4}$ N to the left.

Answer the questions that follow.
7.3.3 Calculate the magnitude of charge T.(5)
Show the memo (5 ticks)Memo
1. ✓
  METHOD 1: Formula $F = \frac{k Q _{1} Q _{2}}{r ^{2}}$
  1 mark · Apply Coulomb's law
2. ✓
  METHOD 1: Both single forces $F_{ST} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 9} ) Q _{T}}{( 0 , 02 ) ^{2}}$ and $F_{PT} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 9} ) Q _{T}}{( 0 , 06 ) ^{2}}$
  1 mark · Apply Coulomb's law
3. ✓
  METHOD 1: Substitute $F_{n e t}$ $F_{n e t} = F_{ST} - F_{PT}$
  $2, 5 \times 1 0^{- 4} = \dots$
  1 mark · Find net electrostatic force or field
4. ✓
  METHOD 1: Subtraction $2, 5 \times 1 0^{- 4} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 9} ) Q _{T}}{( 0 , 02 ) ^{2}} - \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 9} ) Q _{T}}{( 0 , 06 ) ^{2}}$
  1 mark · Find net electrostatic force or field
5. ✓
  METHOD 1: Answer $40000 Q_{T} = 2, 5 \times 1 0^{- 4}$
  $Q_{T} = 6, 25 \times 1 0^{- 9}$ C
  1 mark · Find net electrostatic force or field
4. Question 7.3 · November 2023 · Paper 1 · National (DBE) · 5 marks
Two point charges, $A$ and $B$ , each with a charge of $+ 3 \times 1 0^{- 9}$ C, are stationary on a horizontal surface. Point $P$ is r metres from charge $A$ and 2r metres from charge $B$ , as shown in the diagram below.
The magnitude of the net electric field at point $P$ is $27$ N·C $^{- 1}$ .
7.3 Calculate the value of r.(5)
Show the memo (5 ticks)Memo
1. ✓
  Formula $E = \frac{k Q}{r ^{2}}$
  1 mark · Calculate electric field strength
2. ✓
  Substitution for A or B (any one) $E_{A} = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{r ^{2}}$
  $E_{B} = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{( 2 r ) ^{2}}$
  1 mark · Calculate electric field strength
3. ✓
  Substitution of 27 for Enet $E_{n e t} = E_{A} - E_{B}$
  $27 =$
  1 mark · Find net electrostatic force or field
4. ✓
  Subtraction of the fields $\frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{r ^{2}} - \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{4 r ^{2}}$
  1 mark · Find net electrostatic force or field
5. ✓
  Answer $27 = \frac{27}{r ^{2}} - \frac{6 , 75}{r ^{2}} = \frac{20 , 25}{r ^{2}}$
  $r = 0, 87$ (m)
  1 mark · Find net electrostatic force or field
5. Question 7.6 · November 2022 · Paper 1 · National (DBE) · 5 marks
A charged sphere M is suspended from a ceiling by a light inextensible, insulated string.

Another charged sphere N, of mass $2, 04 \times 1 0^{- 3}$ kg and carrying a charge of $+ 8, 6 \times 1 0^{- 8}$ C, hangs STATIONARY vertically below sphere M. The centres of the spheres are $0, 3$ m apart, as shown in the diagram below.
Point X is $0, 1$ m vertically below the centre of sphere N, as shown below.
7.6 Calculate the net electric field at point X.(5)
Show the memo (5 ticks)Memo
1. ✓
  Formula for EUpwards as positive:
  $E = \frac{k Q}{r ^{2}}$
  1 mark · Calculate electric field strength
2. ✓
  Substitution for M or N $E_{M} = \frac{( 9 \times 1 0 ^{9} ) ( 2 , 33 \times 1 0 ^{- 6} )}{( 0 , 4 ) ^{2}}$ , $E_{N} = \frac{( 9 \times 1 0 ^{9} ) ( 8 , 6 \times 1 0 ^{- 8} )}{( 0 , 1 ) ^{2}}$
  1 mark · Calculate electric field strength
3. ✓
  Subtraction $E_{n e t} = E_{M} - E_{N}$
  $E_{n e t} = 131062, 5 - 77400$
  1 mark · Find net electrostatic force or field
4. ✓
  Answer $E_{n e t} = 53662, 5$ N·C $^{- 1}$ ( $5, 36 \times 1 0^{4}$ N·C $^{- 1}$ )
  1 mark · Find net electrostatic force or field
5. ✓
  Directionupwards / towards M
  1 mark · Find net electrostatic force or field
6. Question 7.4 · November 2025 · Paper 1 · National (DBE) · 6 marks
A sphere X is placed at point A on a horizontal surface. X carries a charge of $+ 3 \times 1 0^{- 7}$ C. Point P is r metres to the left of point A, as shown in the diagram below.

The magnitude of the electric field at point P is $1, 08 \times 1 0^{6}$ N·C $^{- 1}$ .
Sphere Y, carrying a charge of $- 5 \times 1 0^{- 7}$ C, is now fixed at point P and sphere Z, carrying a charge of $- 4 \times 1 0^{- 7}$ C, is fixed $0, 04$ m to the right of sphere X, as shown in the diagram below.
7.4 The NET FORCE acting on sphere X is $0, 0427$ N at point A.

Is the surface frictionless? Choose from YES or NO. Explain the answer by means of a calculation.(6)
Show the memo (6 ticks)Memo
1. ✓
  METHOD 1: Formula $F_{X Y} = \frac{k Q _{X} Q _{Y}}{r ^{2}}$
  1 mark · Apply Coulomb's law
2. ✓
  METHOD 1: Force of Y on X $F_{X Y} = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 7} ) ( 5 \times 1 0 ^{- 7} )}{( 0 , 05 ) ^{2}} = 0, 54$ N
  1 mark · Apply Coulomb's law
3. ✓
  METHOD 1: Force of Z on X $F_{XZ} = \frac{k Q _{X} Q _{Z}}{r ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 7} ) ( 4 \times 1 0 ^{- 7} )}{( 0 , 04 ) ^{2}} = 0, 68$ N
  1 mark · Apply Coulomb's law
4. ✓
  METHOD 1: Net electrostatic force $F_{n e t (E)} = F_{XZ} + F_{X Y}$
  $= 0, 68 - 0, 54 = 0, 14$ N
  1 mark · Find net electrostatic force or field
5. ✓
  METHOD 1: ChoiceNo
  1 mark · Find net electrostatic force or field
6. ✓
  METHOD 1: Reason $F_{n e t (E)} > F_{n e t}$ (or $F_{n e t (E)} \neq = F_{n e t}$ )
  1 mark · Find net electrostatic force or field

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