PhysicsCBSE Class 12 · Electric Charges and Fields

Coulomb's Law and Superposition Class 12: Forces Between Charges

Q: Does Coulomb's law work for charged spheres?

Yes, as long as the spheres are small compared with the distance between them (they behave as point charges). For a uniformly charged spherical shell, the field outside acts as if all the charge were at the centre, so you can use the centre-to-centre distance.

Q: Why is Coulomb's law called an inverse square law?

Because the force varies as 1/r². Doubling the distance makes the force one-quarter, tripling it makes the force one-ninth.

Q: What is the difference between Coulomb's law and the superposition principle?

Coulomb's law gives the force between one pair of charges. The superposition principle says how to combine several such forces: add them as vectors, each found as if the other charges were not there.

Q: Is Coulomb's force always greater than gravitational force?

Between elementary particles, yes, by a huge factor: for an electron and a proton the electric force is about 10³⁹ times the gravitational force. Large bodies are nearly neutral, so gravity dominates on the scale of planets.

Q: Why does Coulomb's law obey Newton's third law?

The vector form gives \vec F_{12} = -\vec F_{21}: the forces on the two charges are equal in size and opposite in direction, acting along the same line.

Updated 15 October 2026 · Fact-checked

Coulomb's law says two point charges at rest push or pull each other with a force proportional to the product of the charges and inversely proportional to the square of their separation, along the line joining them. With several charges, the superposition principle adds the separate forces as vectors to give the net force.

Each charge pushes q2 along the line joining them, with F = kq₁q₂/r²; add these forces as vectors, tip to tail, to get the net force.

Illustration drawn for QuestionBuddy.

Understand Coulomb's Law and Superposition

Rub a plastic comb on dry hair and it lifts tiny bits of paper. The pull is strong enough to beat the weight of the paper, and it gets weaker quickly as you move the comb away. Coulomb measured exactly how this force depends on the charges and the distance between them.

Coulomb's law. Two point charges $q_{1}$ and $q_{2}$ at rest, a distance $r$ apart in vacuum, exert on each other a force of magnitude $F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}}$ , where $k = \frac{1}{4 π ε _{0}} \approx 9 \times 1 0^{9} N m^{2} C^{- 2}$ and $ε_{0} = 8.854 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ is the permittivity of free space. The force acts along the line joining the charges: like charges repel, unlike charges attract. A 'point charge' means the size of the charged body is tiny compared with the distance between the bodies.

Vector form. Let $r_{21} = r_{2} - r_{1}$ point from $q_{1}$ to $q_{2}$ . The force on $q_{2}$ due to $q_{1}$ is $F_{21} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{21}^{2}} \overset{r}{^}_{21}$ . Put the signs of the charges in: if $q_{1} q_{2} > 0$ the force on $q_{2}$ points away from $q_{1}$ (repulsion); if $q_{1} q_{2} < 0$ it points towards $q_{1}$ (attraction). Swapping the labels gives $F_{12} = - F_{21}$ , so Coulomb's law obeys Newton's third law.

Unit of charge. One coulomb is the charge that, placed 1 m from an equal charge in vacuum, repels it with a force of $9 \times 1 0^{9}$ N. That is an enormous force, which tells you 1 C is a very large charge; lab charges are in $μ$ C or nC. Charge is also quantised ( $q = n e$ , with $e = 1.6 \times 1 0^{- 19}$ C) and conserved (it is only transferred, never created alone).

In a medium. If the charges sit in a medium of permittivity $ε = K ε_{0}$ , the force becomes $F_{m} = \frac{1}{4 π ε _{0} K} \frac{∣ q _{1} q _{2} ∣}{r ^{2}} = \frac{F _{v a c uu m}}{K}$ . $K$ (the dielectric constant or relative permittivity) is greater than 1 for every material, so a medium always weakens the force.

Superposition principle. When many charges are present, the force between any two of them is exactly what Coulomb's law gives, unaffected by the others. So the net force on $q_{1}$ is the vector sum $F_{1} = F_{12} + F_{13} + \dots + F_{1 n}$ . In practice: find each force's size with $k ∣ q_{i} q_{j} ∣/ r^{2}$ , mark its direction on a diagram (towards or away from the other charge), then add the vectors by components or by the parallelogram law, using symmetry wherever it cancels components.

From force to field. Dividing the force on a small test charge by its charge gives the electric field, $E = \frac{F}{q _{0}}$ . For a point charge $Q$ , $E = \frac{k Q}{r ^{2}}$ , pointing away from a positive $Q$ and towards a negative one. Fields add by superposition in exactly the same way as forces. A neat consequence: between two like charges there is a point where the field is zero (nearer the smaller charge); for two unlike charges the zero-field point lies outside the pair, on the side of the smaller charge.

Continuous charge. When charge is spread over a line, a surface or a volume, describe it by $λ$ (C/m), $σ$ (C/m²) or $ρ$ (C/m³), split it into tiny pieces $d q$ that act as point charges, and add their effects. You will meet this idea again in Gauss's law.

Coulomb's Law and Superposition: formulas and key facts

Coulomb's law (magnitude): $F = \frac{1}{4 π ε _{0}} \frac{∣ q _{1} q _{2} ∣}{r ^{2}}$; $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}$ . Use SI units: C and m.
Coulomb's law (vector form): $F_{21} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{21}^{2}} \overset{r}{^}_{21}, F_{12} = - F_{21}$; $\overset{r}{^}_{21}$ points from $q_{1}$ to $q_{2}$ . Keep the signs of the charges in the vector form.
Force in a medium: $F_{m} = \frac{F _{0}}{K}, ε = K ε_{0}$; K is the dielectric constant (relative permittivity) of the medium; K = 1 for vacuum.
Superposition of forces: $F_{1} = F_{12} + F_{13} + \dots + F_{1 n}$; Add as vectors; equal forces at angle $θ$ give $2 F cos \frac{θ}{2}$ .
Field of a point charge: $E = \frac{1}{4 π ε _{0}} \frac{∣ Q ∣}{r ^{2}}, F = q E$; Away from a positive charge, towards a negative charge.
Quantisation of charge: $q = n e, e = 1.6 \times 1 0^{- 19} C$; n is a whole number; 1 C needs about $6.25 \times 1 0^{18}$ electrons.
Charge densities: $λ = \frac{d q}{d l}, σ = \frac{d q}{d A}, ρ = \frac{d q}{d V}$; Used for continuous charge distributions.

How to solve Coulomb's Law and Superposition questions

Use this method for any question on forces or fields due to two or more point charges.

1Draw the charges with their signs and positions; convert μC to $1 0^{- 6}$ C, nC to $1 0^{- 9}$ C and cm to m.
2Pick the charge on which the force is wanted. For each other charge, find the magnitude $k ∣ q_{i} q_{j} ∣/ r^{2}$ (no signs yet).
3Mark each force's direction on the diagram: away from a like charge, towards an unlike charge.
4Add the forces as vectors: along a line, add with + and − signs; at an angle, use components or $F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos θ$ .
5Look for symmetry first: equal forces at equal angles cancel sideways and add along the bisector.
6For equilibrium or zero-field questions, set the two opposing forces (or fields) equal and solve for the distance; check the point lies where the directions really are opposite.
7Write the answer with magnitude, unit and direction (angle or axis).

Coulomb's Law and Superposition in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Coulomb force balancing gravity

Suppose you hang two identical small pith balls, each of mass 0.1 g, from the same hook with silk threads 50 cm long, and give them equal charges by touching them with a charged rod. They spread apart and come to rest 6 cm apart. Take g = 9.8 m/s².

(a) Find the electric force between the balls. (b) Find the charge on each ball. (c) If the charge on each ball were doubled, by what factor would the force at the same separation change, and would the balls move closer or further apart?

Try it first, then show the solutionSolution

Each thread makes angle θ with the vertical, with $sin θ = \frac{3 cm}{50 cm} = 0.06$ , so $tan θ \approx 0.0601$ .
Each ball is in equilibrium under weight $m g$ , tension and the electric force $F$ , so $F = m g tan θ = 1 0^{- 4} \times 9.8 \times 0.0601 \approx 5.89 \times 1 0^{- 5}$ N.
$F = \frac{k q ^{2}}{r ^{2}}$ gives $q = r \frac{F}{k} = 0.06 \frac{5.89 \times 1 0 ^{- 5}}{9 \times 1 0 ^{9}} \approx 4.85 \times 1 0^{- 9}$ C.
Doubling both charges multiplies $q_{1} q_{2}$ by 4, so the force at 6 cm becomes 4 times larger; this is more than the new $m g tan θ$ can balance, so the balls swing further apart until a new balance is reached.

Answer: (a) about $5.9 \times 1 0^{- 5}$ N; (b) about 4.9 nC on each ball; (c) the force becomes 4 times larger and the balls move further apart.

Memory hook: Two hanging balls: tan θ = F/mg turns an angle into a charge.

How big one coulomb is

Imagine you could put a charge of 1 C on each of two small spheres and hold them 1 km apart in air.

(a) Find the force between them. (b) What mass would have this much weight on Earth (g = 9.8 m/s²)? (c) Why do lab experiments use microcoulombs instead?

Try it first, then show the solutionSolution

$F = \frac{9 \times 1 0 ^{9} \times 1 \times 1}{( 1000 ) ^{2}} = 9 \times 1 0^{3}$ N.
Mass with this weight: $m = \frac{F}{g} = \frac{9000}{9.8} \approx 918$ kg.
Even 1 km apart, 1 C charges push with the weight of a small car, so 1 C is huge; everyday charging by rubbing gives only nC to μC.

Answer: (a) $9 \times 1 0^{3}$ N; (b) about 918 kg; (c) because 1 C produces enormous forces; practical charges are millionths of a coulomb or less.

Memory hook: 1 C at 1 km weighs as much as a small car: the coulomb is a giant unit.

Tricks to remember Coulomb's Law and Superposition

Short memory aids. Say each one aloud once, then use it on a question the same day.

Inverse square, square the distance
Double the distance, quarter the force; triple it, one-ninth. Change the charges and the force changes in the same ratio as their product.
Use it when: Ratio questions where r and q both change.
Like push, unlike pull, always along the line
Draw the arrow on the line joining the two charges: away for like charges, towards for unlike.
Use it when: Getting directions right before adding vectors.
Zero field: between for like, outside for unlike
Like charges: the null point is between them, nearer the smaller one. Unlike charges: outside, beyond the smaller one.
Use it when: Locating the point of zero field or zero force quickly.
Equal forces at an angle
Two equal forces F with angle θ between them add to $2 F cos (θ /2)$ along the bisector. For an equilateral triangle (60°) that is $3 F$ .
Use it when: Triangle and square arrangements.

Coulomb's Law and Superposition: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Force in air and in a medium · 2 marks · Very short answerBoard level

Two point charges of +4 μC and −6 μC are 30 cm apart in air.
(a) Find the magnitude and nature of the force between them. (1)
(b) The charges are placed, with the same separation, in a medium of dielectric constant 4. Find the new force. (1)

Try it first, then show the solutionSolution

(a) $F = \frac{9 \times 1 0 ^{9} \times 4 \times 1 0 ^{- 6} \times 6 \times 1 0 ^{- 6}}{( 0.3 ) ^{2}} = \frac{0.216}{0.09} = 2.4$ N. The charges are unlike, so the force is attractive. (1)
(b) In a medium $F_{m} = \frac{F}{K} = \frac{2.4}{4} = 0.6$ N, still attractive. (1)

Answer: (a) 2.4 N, attractive; (b) 0.6 N.

Example 2: Charges at the corners of an equilateral triangle · 3 marks · Short answerHarder board level

Three equal charges of +2 μC each are placed at the corners A, B and C of an equilateral triangle of side 10 cm. Find the magnitude and direction of the net force on the charge at C.

Try it first, then show the solutionSolution

Force on C due to A, and on C due to B, each has magnitude $F = \frac{9 \times 1 0 ^{9} \times ( 2 \times 1 0 ^{- 6} ) ^{2}}{( 0.1 ) ^{2}} = 3.6$ N. (1)
Both are repulsive: one along AC (away from A) and one along BC (away from B). The angle between them is 60°. (1)
Components perpendicular to the bisector of angle C cancel; along the bisector, $F_{n e t} = 2 F cos 30° = 3 \times 3.6 \approx 6.24$ N, pointing away from the midpoint of AB. (1)

Answer: About 6.24 N along the bisector of angle C, directed away from side AB.

Example 3: Equilibrium of three charges · 5 marks · Long answerHOTS / competency

A charge +q is fixed at x = 0 and +4q at x = L.
(a) Where on the line should a third charge Q be placed so that the net force on Q is zero? (2)
(b) Find Q (sign and size) so that all three charges are in equilibrium. (2)
(c) Is the equilibrium of Q in (a) stable for small displacements along the line if Q is positive? Give a reason. (1)

Try it first, then show the solutionSolution

(a) Q must be between the charges (only there do the two forces on it point in opposite directions). At distance x from +q: $\frac{k q Q}{x ^{2}} = \frac{k ( 4 q ) Q}{( L - x ) ^{2}}$ . (1)
So $(L - x)^{2} = 4 x^{2}$ , $L - x = 2 x$ , $x = \frac{L}{3}$ . This does not depend on Q. (1)
(b) For +q to be in equilibrium, the pull of Q (at $L /3$ ) must balance the push of +4q (at L), so Q must be negative: $\frac{k q ∣ Q ∣}{( L /3 ) ^{2}} = \frac{k q ( 4 q )}{L ^{2}}$ . (1)
$9∣ Q ∣ = 4 q$ , so $Q = - \frac{4 q}{9}$ . Check on +4q: pull of Q $= \frac{k ( 4 q ) ( 4 q /9 )}{( 2 L /3 ) ^{2}} = \frac{4 k q ^{2}}{L ^{2}}$ , push of +q $= \frac{4 k q ^{2}}{L ^{2}}$ ; they balance. (1)
(c) For positive Q moved slightly towards +q, the push from +q grows faster than that from +4q falls, so Q is pushed back: stable along the line. (1)

Answer: (a) At x = L/3 from +q; (b) Q = −4q/9; (c) stable for displacements along the line, since the nearer charge pushes Q back.

Example 4: Touching identical spheres · 4 marks · Case-basedHarder board level

Read the passage and answer.

Two identical small metal spheres carry charges of +6 μC and −2 μC and are held 20 cm apart in air. They are touched together and put back at the same separation.
(a) Find the force between them before touching. (1)
(b) Find the charge on each sphere after touching, and the law used. (1)
(c) Find the new force and its nature. (1)
(d) Find the ratio of the new force to the old one. (1)

Try it first, then show the solutionSolution

(a) $F = \frac{9 \times 1 0 ^{9} \times 6 \times 1 0 ^{- 6} \times 2 \times 1 0 ^{- 6}}{( 0.2 ) ^{2}} = 2.7$ N, attractive. (1)
(b) Total charge is conserved: $+ 6 - 2 = + 4$ μC, shared equally by identical spheres, so +2 μC each. (1)
(c) $F^{'} = \frac{9 \times 1 0 ^{9} \times ( 2 \times 1 0 ^{- 6} ) ^{2}}{( 0.2 ) ^{2}} = 0.9$ N, now repulsive. (1)
(d) $\frac{F ^{'}}{F} = \frac{0.9}{2.7} = \frac{1}{3}$ . (1)

Answer: (a) 2.7 N attractive; (b) +2 μC each (conservation of charge); (c) 0.9 N repulsive; (d) 1 : 3.

Example 5: Field at the centre of a square · 3 marks · Short answerHarder board level

Charges +q, +q, −q and −q are placed at the corners A, B, C and D (in order) of a square of side a. Find the electric field at the centre O. Evaluate it for q = 1 μC and a = 10 cm.

Try it first, then show the solutionSolution

Each corner is $\frac{a}{2}$ from O, so each charge alone gives $E_{0} = \frac{k q}{a ^{2} /2} = \frac{2 k q}{a ^{2}}$ . (1)
Field of +q at A points from A towards O, which is the same direction as the field of −q at C (towards C). Similarly, the field of +q at B and of −q at D both point from B towards D. So the two diagonal pairs give $2 E_{0}$ each, at 90° to each other. (1)
Resultant $= (2 E_{0})^{2} + (2 E_{0})^{2} = 22 E_{0} = \frac{4 2 k q}{a ^{2}}$ , perpendicular to AB, pointing from side AB towards side CD. Numerically $\frac{4 2 \times 9 \times 1 0 ^{9} \times 1 0 ^{- 6}}{0.01} \approx 5.09 \times 1 0^{6}$ N/C. (1)

Answer: $E = \frac{4 2 k q}{a ^{2}} \approx 5.09 \times 1 0^{6}$ N/C, perpendicular to AB, from the positive side towards the negative side.

Example 6: Coulomb's law in vector form · 3 marks · Short answerHarder board level

A charge $q_{1} = + 1$ μC is at the origin and $q_{2} = - 2$ μC is at the point (3 cm, 4 cm, 0). Find the force on $q_{2}$ due to $q_{1}$ in vector form and show that $F_{12} = - F_{21}$ .

Try it first, then show the solutionSolution

$r_{21} = (0.03 \hat{i} + 0.04 \hat{j})$ m, $r = 0.05$ m, $\overset{r}{^}_{21} = 0.6 \hat{i} + 0.8 \hat{j}$ . (1)
$F_{21} = \frac{k q _{1} q _{2}}{r ^{2}} \overset{r}{^}_{21} = \frac{9 \times 1 0 ^{9} \times ( 1 0 ^{- 6} ) ( - 2 \times 1 0 ^{- 6} )}{0.0025} (0.6 \hat{i} + 0.8 \hat{j}) = - 7.2 (0.6 \hat{i} + 0.8 \hat{j}) = (- 4.32 \hat{i} - 5.76 \hat{j})$ N. (1)
The minus sign means the force on $q_{2}$ points towards $q_{1}$ (attraction). For the force on $q_{1}$ , $\overset{r}{^}_{12} = - \overset{r}{^}_{21}$ , so $F_{12} = (4.32 \hat{i} + 5.76 \hat{j})$ N $= - F_{21}$ . (1)

Answer: $F_{21} = (- 4.32 \hat{i} - 5.76 \hat{j})$ N, magnitude 7.2 N towards $q_{1}$ ; $F_{12}$ is equal and opposite.

Where students lose marks in Coulomb's Law and Superposition

The slips examiners see most often, and what to write instead.

Mistake: Putting signs of charges into the magnitude formula and then also guessing the direction.
A negative answer for a magnitude is then 'corrected' by flipping the arrow, giving the wrong direction.
Wrong working
F = k(4 μC)(−6 μC)/r² = −2.4 N, so the force is 2.4 N repulsive.
Right working
|F| = k(4 μC)(6 μC)/r² = 2.4 N; unlike charges, so attractive.
Do this: Use magnitudes for size and decide direction from like/unlike; keep signs only in the vector form.
Usually costs: ½ to 1 mark for the wrong nature of force.
Mistake: Adding forces from different charges as plain numbers.
Forces at an angle cannot be added arithmetically.
Wrong working
Net force on C = 3.6 + 3.6 = 7.2 N.
Right working
Net force on C = 2 × 3.6 × cos 30° ≈ 6.24 N along the bisector.
Do this: Draw the arrows first; add by components or the parallelogram law.
Usually costs: 1 to 2 marks in a 3-mark superposition question.
Mistake: Leaving distances in cm or charges in μC.
A 10 cm distance put in as 10 makes the force $1 0^{4}$ times too small.
Wrong working
F = 9 × 10⁹ × 4 × 6/(30)²
Right working
F = 9 × 10⁹ × (4 × 10⁻⁶)(6 × 10⁻⁶)/(0.3)²
Do this: Convert every value to C and m in the first line of working.
Usually costs: The final answer mark.
Mistake: Placing the zero-field point between two unlike charges.
Between unlike charges both fields point the same way and can never cancel.
Do this: Check the directions of the two fields in each region before solving the equation.
Usually costs: Full marks of a 2 to 3 mark question.
Mistake: Forgetting that the dielectric constant reduces the force.
Students multiply by K instead of dividing.
Wrong working
F in medium = 2.4 × 4 = 9.6 N
Right working
F in medium = 2.4/4 = 0.6 N
Do this: Remember $ε = K ε_{0}$ appears in the denominator.
Usually costs: 1 mark.

Exam tips

When asked to state Coulomb's law, mention point charges, at rest, product of charges, inverse square of distance, and that the force acts along the line joining them.
Write the vector form with $\overset{r}{^}_{21}$ defined in words, and add one line showing $F_{12} = - F_{21}$ when a 2-mark question asks for it.
In superposition questions, a neat labelled diagram with every force arrow earns a step mark even if the arithmetic slips.
For the 'how many electrons' type question, state whether electrons were gained or lost, not just the number.
Board papers of 2024 (55/5/1, Q17) and 2025 (55/1/1, Q1) set short questions on superposition and on the F versus 1/r² graph; the slope of that graph is $k q_{1} q_{2}$ , so a steeper line means a larger product of charges.

Practice questions on Coulomb's Law and Superposition

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The force between two point charges is F. One charge is doubled and the distance between them is tripled. What is the new force?
Hint
F ∝ q₁q₂/r².
Answer
$F^{'} = F \times \frac{2}{3 ^{2}} = \frac{2 F}{9}$ .
Q2 · 1 mark · Very short answerBoard level
Write the SI unit of the permittivity of free space $ε_{0}$ , and state the value of $\frac{1}{4 π ε _{0}}$ .
Hint
Rearrange $F = \frac{q _{1} q _{2}}{4 π ε _{0} r ^{2}}$ for $ε_{0}$ .
Answer
$ε_{0} = \frac{q _{1} q _{2}}{4 π F r ^{2}}$ , so its unit is C² N⁻¹ m⁻²; $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9}$ N m² C⁻².
Q3 · 2 marks · Very short answerBoard level
A body is given a charge of $- 3.2 \times 1 0^{- 9}$ C by rubbing. (a) Did it gain or lose electrons, and how many? (b) Which property of charge does your answer use?
Hint
n = q/e.
Answer
(a) It gained electrons: $n = \frac{3.2 \times 1 0 ^{- 9}}{1.6 \times 1 0 ^{- 19}} = 2 \times 1 0^{10}$ electrons. (b) Quantisation of charge ( $q = n e$ ).
Q4 · 2 marks · Very short answerHarder board level
Charges +1 μC, −2 μC and +4 μC are placed at x = 0, x = 10 cm and x = 20 cm on the x-axis. Find the net force on the −2 μC charge.
Hint
Both charges attract the middle charge, in opposite directions.
Answer
Pull towards +1 μC: $\frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 12}}{0.01} = 1.8$ N along −x. Pull towards +4 μC: $\frac{9 \times 1 0 ^{9} \times 8 \times 1 0 ^{- 12}}{0.01} = 7.2$ N along +x. Net force = 5.4 N along +x (towards the +4 μC charge).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The force between two point charges is not changed when a third charge is brought close to them.
Reason (R): Coulomb's law gives the force between two charges independently of other charges present, and the net force on a charge is the vector sum of these separate forces.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Separate 'force between two charges' from 'net force on one charge'.
Answer
(a). The mutual force of the pair is unchanged (superposition principle); only the net force on each charge changes because the third charge adds its own force.
Q6 · 3 marks · Short answerHOTS / competency
Charges of +2 μC and −8 μC are 30 cm apart. Find the point on the line joining them where the electric field is zero.
Hint
For unlike charges the null point lies outside, on the side of the smaller charge.
Answer
Let the point be d from the +2 μC charge, on the side away from −8 μC: $\frac{2}{d ^{2}} = \frac{8}{( d + 30 ) ^{2}}$ gives $d + 30 = 2 d$ , so d = 30 cm. The field is zero 30 cm from the +2 μC charge, on the side opposite the −8 μC charge (60 cm from the −8 μC charge).
Q7 · 3 marks · Short answerHOTS / competency
Two identical metal spheres A and B carry equal charges q and repel each other with force F. A third identical uncharged sphere C touches A, then touches B, and is then removed. Find the new force between A and B at the same separation.
Hint
Share the charge equally at each touch.
Answer
After touching A: A and C have $\frac{q}{2}$ each. After touching B: B and C share $q + \frac{q}{2}$ , so B has $\frac{3 q}{4}$ . New force $F^{'} = F \times \frac{( q /2 ) ( 3 q /4 )}{q ^{2}} = \frac{3 F}{8}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Coulomb's law in one line.Show answer
The force between two point charges at rest is proportional to the product of the charges and inversely proportional to the square of the distance between them, acting along the line joining them.
2.What is the value of $\frac{1}{4 π ε _{0}}$ ?Show answer
$9 \times 1 0^{9}$ N m² C⁻².
3.How does the force between two charges change in a medium of dielectric constant K?Show answer
It becomes 1/K times the force in vacuum.
4.State the superposition principle for electric forces.Show answer
The net force on a charge is the vector sum of the forces due to each other charge, each calculated as if the others were absent.
5.Where is the zero-field point for two like charges?Show answer
Between them, nearer the smaller charge.
6.If the distance between two charges is halved, what happens to the force?Show answer
It becomes 4 times larger.

Coulomb's Law and Superposition: frequently asked questions

Does Coulomb's law work for charged spheres?

Yes, as long as the spheres are small compared with the distance between them (they behave as point charges). For a uniformly charged spherical shell, the field outside acts as if all the charge were at the centre, so you can use the centre-to-centre distance.

Why is Coulomb's law called an inverse square law?

Because the force varies as 1/r². Doubling the distance makes the force one-quarter, tripling it makes the force one-ninth.

What is the difference between Coulomb's law and the superposition principle?

Coulomb's law gives the force between one pair of charges. The superposition principle says how to combine several such forces: add them as vectors, each found as if the other charges were not there.

Is Coulomb's force always greater than gravitational force?

Between elementary particles, yes, by a huge factor: for an electron and a proton the electric force is about 10³⁹ times the gravitational force. Large bodies are nearly neutral, so gravity dominates on the scale of planets.

Why does Coulomb's law obey Newton's third law?

The vector form gives \vec F_{12} = -\vec F_{21}: the forces on the two charges are equal in size and opposite in direction, acting along the same line.