Coulomb’s law & electric fields: practice problems with step-by-step solutions

Free Coulomb’s law & electric fields practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: force between two charges, field between two charges, forces at a right angle.

Free practice problemEasy
+ + + + + +− − − − − −e⁻E = 3300 N/C

An electron (charge −1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg) is in a uniform electric field of 3300 N/C. Ignoring gravity, find the magnitude of its acceleration a.

E = 3300 N/C|q| = 1.602 × 10⁻¹⁹ Cm = 9.109 × 10⁻³¹ kg
m/s²
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What you practise

  1. Easyforce between two charges
  2. Mediumfield between two charges
  3. Hardforces at a right angle

Key formulas

Coulomb’s law
F = k·|q1q2|/r²
Field of a point charge
E = k·|q|/r²
Constant
k = 8.99 × 10⁹ N·m²/C²
Force on a charge
F = q·E

Learn Coulomb’s law & electric fields from scratch

Before this: Newton’s laws

1What electric charge does

Rub a balloon on your hair and it sticks to the wall. The balloon has picked up electric charge, and charges push or pull on each other without touching.

There are two kinds of charge, positive and negative. Charge q is measured in coulombs (C). One coulomb is huge, so questions use microcoulombs: 1 µC = 10⁻⁶ C, and nanocoulombs: 1 nC = 10⁻⁹ C.

  • Like charges (+ and +, or − and −) repel: they push apart.
  • Unlike charges (+ and −) attract: they pull together.
++FFr+−FF

The two forces are always equal in size and opposite in direction, as Newton’s third law says.

Self-check A charge of +3 µC sits near a charge of −2 µC. What does the force between them do?
  1. It is zero because the charges are different
  2. It only acts on the larger charge
  3. It pulls them together
  4. It pushes them apart

Answer: It pulls them together. A positive and a negative charge are unlike charges, so they attract. Each one feels a force of the same size, towards the other.

2Coulomb’s law

The size of the force between two point charges q₁ and q₂ at a distance r is:

F = k · |q₁ · q₂| / r² k = 8.99 × 10⁹ N·m²/C²

Bigger charges give a bigger force. The force falls with the square of the distance: twice as far gives a quarter of the force.

Use the sizes of the charges in the formula (the |…| means “without sign”). Decide the direction separately: repel or attract.

Always work in SI units: charges in C and distances in m. 5 cm is 0.05 m, and r² is then 0.0025 m².

Self-check Two charges are moved to twice their original distance. What happens to the force between them?
  1. It becomes 4 times smaller
  2. It doubles
  3. It halves
  4. It stays the same

Answer: It becomes 4 times smaller. F is proportional to 1/r². Doubling r divides the force by 2² = 4. Halving it would mean the force fell with r, not r².

3The electric field

A charge changes the space around it. Any other charge placed there feels a force. The electric field E tells you the force per coulomb at that point, in N/C.

E = k · |q| / r² F = q · E

The field has a direction: away from a positive charge and towards a negative charge. It is the direction a small positive test charge would be pushed.

With two or more charges, find the field of each one separately, then add them as arrows. Along one line, arrows in the same direction add and opposite arrows subtract.

At a right angle, use Pythagoras: F = √(F₁² + F₂²).

4The method step by step

  1. Convert to SI: µC × 10⁻⁶ and nC × 10⁻⁹ to C, cm to m.
  2. Draw the charges and the point you are interested in.
  3. Find the size of each force or field with k·|q₁q₂|/r² or k·|q|/r². Use the distance from that charge to the point.
  4. Draw the direction of each one: repel or attract for forces, away from + and towards − for fields.
  5. Add them as arrows: add or subtract along a line, Pythagoras at a right angle.
  6. Check the order of magnitude: forces between a few µC a few cm apart are a few newtons to a few hundred newtons.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A charge of +2 µC and a charge of −2 µC are 20 cm apart. Point P lies halfway between them. Find the magnitude of the electric field E at P. Use k = 8.99 × 10⁹ N·m²/C².

1

SI units

q₁ = 2 × 10⁻⁶ C q₂ = −2 × 10⁻⁶ C r = 0.1 m to each charge
2

Field of the positive charge

E₁ = 8.99 × 10⁹ · 2 × 10⁻⁶ / 0.1² = 1.798 × 10⁶ N/C

It points away from the positive charge, so to the right.

3

Field of the negative charge

Same charge size and same distance, so the same size: E₂ = 1.798 × 10⁶ N/C.

It points towards the negative charge, which is also to the right.

4

Add the fields

+−P+2 µC−2 µCE1​E2​10 cm10 cm

Both arrows point the same way, so they add.

E = 1.798 × 10⁶ + 1.798 × 10⁶ = 3.60 × 10⁶ N/C
5

Check the answer

If both charges were +2 µC, the two fields at P would point in opposite directions and cancel to zero. Here the signs differ, so they add instead.

Answer

E ≈ 3.60 × 10⁶ N/C, pointing towards the negative charge.

6Common mistakes

  • Leaving the distance in cm. The answer is then 10 000 times too small.
  • Forgetting to square the distance.
  • Forgetting the 10⁻⁶ of µC.
  • Adding fields that point in opposite directions. Draw the arrows first.
  • Using the distance between the two charges instead of the distance from each charge to the point P.
Self-check Charges of 2 µC and 5 µC are 10 cm apart. What is the force between them? Use k = 8.99 × 10⁹ N·m²/C².
  1. 0.899 N
  2. 8.99 N
  3. 8.99 × 10¹² N
  4. 8.99 × 10⁻⁴ N

Answer: 8.99 N. F = 8.99 × 10⁹ · 2 × 10⁻⁶ · 5 × 10⁻⁶ / 0.1² = 8.99 N. Using r = 10 (cm) gives 8.99 × 10⁻⁴ N. Not squaring 0.1 gives 0.899 N. Leaving out 10⁻⁶ gives 8.99 × 10¹² N.

7How exam questions look

  • Easy: the force between two charges with Coulomb’s law, the distance for a given force, the field of one charge, or how the force changes when charges or distance change.
  • Medium: the field between two charges, the point where the field is zero, the net force on a charge between two others, or spheres that share charge after touching.
  • Hard: forces at a right angle, the field on the perpendicular bisector, charged balls hanging on threads, or an electron deflected between plates.

For “what if” questions, use ratios: tripling one charge and halving the distance multiplies the force by 3/0.5² = 12.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
++4 µC++1 µCr = ?

Point charges of +4 µC and +1 µC exert a force of 20 N on each other. How far apart r are they? Give the answer in cm. Use k = 8.99 × 10⁹ N·m²/C².

q1 = +4 µCq2 = +1 µCF = 20 N

Find r (cm)

Show hints
  1. Rearrange Coulomb’s law for r.
  2. r = √(k·|q₁·q₂|/F).
  3. r² = 0.001798 m².
Show the answer and the step-by-step solution

Coulomb’s law: two point charges attract or repel with F = k·|q₁·q₂|/r², with k = 8.99 × 10⁹ N·m²/C². The field of a point charge is E = k·|q|/r², pointing away from a positive charge and towards a negative one. Forces and fields from several charges add as vectors.

Solve for the distance

r = √(8.99 × 10⁹·4 × 10⁻⁶·1 × 10⁻⁶/20) = 0.0424 m
r = 4.24 cm

Answer

r = 4.24 cm
Example 2Medium
++10 µC−−6 µCr = 40 cm

Two identical small metal spheres carry charges of +10 µC and −6 µC. They are touched together and then put back 40 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².

q1 = +10 µCq2 = −6 µCr = 40 cm

Find F (N)

Show hints
  1. Touching identical conductors share the total charge equally.
  2. Each sphere ends with (q₁ + q₂)/2.
  3. q′ = +2 µC.
Show the answer and the step-by-step solution

Coulomb’s law: two point charges attract or repel with F = k·|q₁·q₂|/r², with k = 8.99 × 10⁹ N·m²/C². The field of a point charge is E = k·|q|/r², pointing away from a positive charge and towards a negative one. Forces and fields from several charges add as vectors.

Charge after touching

q′ = (10 + −6)/2 = 2 µC

New force

F = 8.99 × 10⁹·(2 × 10⁻⁶)²/0.4² = 0.2247 N

Answer

F = 0.2247 N

Both spheres now carry the same sign, so they repel.

Example 3Hard
L = 50 cm5°qq

Two identical small balls, each of mass 5 g, hang from the same point on insulating threads 50 cm long. They are given equal charges and each thread settles at 5° from the vertical. Use g = 9.81 m/s². Find the charge q on each ball in nC. Use k = 8.99 × 10⁹ N·m²/C².

m = 5 gL = 50 cmθ = 5°g = 9.81 m/s²

Find q (nC)

Show hints
  1. Each ball is in equilibrium under gravity, the thread tension and the electric repulsion.
  2. tan θ = F/(m·g), and the separation is r = 2·L·sin θ.
  3. F = 0.004291 N, r = 0.08716 m.
Show the answer and the step-by-step solution

Coulomb’s law: two point charges attract or repel with F = k·|q₁·q₂|/r², with k = 8.99 × 10⁹ N·m²/C². The field of a point charge is E = k·|q|/r², pointing away from a positive charge and towards a negative one. Forces and fields from several charges add as vectors.

Electric force from the equilibrium

The tension's horizontal part balances F, its vertical part balances m·g.

F = m·g·tan θ = 0.005·9.81·tan 5° = 0.004291 N

Separation

r = 2·L·sin θ = 2·0.5·sin 5° = 0.08716 m

Charge

F = k·q²/r² → q = r·√(F/k) = 0.08716·√(0.004291/8.99 × 10⁹) = 6.022 × 10⁻⁸ C = 60.22 nC

Answer

q = 60.22 nC

Questions and answers

What is Coulomb’s law & electric fields?

Rub a balloon on your hair and it sticks to the wall. The balloon has picked up electric charge, and charges push or pull on each other without touching.

There are two kinds of charge, positive and negative. Charge q is measured in coulombs (C). One coulomb is huge, so questions use microcoulombs: 1 µC = 10⁻⁶ C, and nanocoulombs: 1 nC = 10⁻⁹ C.

How do I solve Coulomb’s law & electric fields problems step by step?

Coulomb’s law: two point charges attract or repel with F = k·|q₁·q₂|/r², with k = 8.99 × 10⁹ N·m²/C². The field of a point charge is E = k·|q|/r², pointing away from a positive charge and towards a negative one. Forces and fields from several charges add as vectors.

What are common mistakes in Coulomb’s law & electric fields?
  • Take the square root: r = √(k·|q₁·q₂|/F).
  • That is the distance in metres. Convert to cm.
  • That is the force before touching. Identical spheres share the total charge equally.
What do the Coulomb’s law & electric fields exercises cover?
  • Easy: force between two charges
  • Medium: field between two charges
  • Hard: forces at a right angle

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

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Newton’s laws

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Magnetic fields & induction

Calculators

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Courses with this topic

Physics 2: Electricity and Magnetism

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Coulomb’s law & electric fields worksheet with answers
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