The big idea: There are two kinds of electric charge: positive and negative (unit: the coulomb, C).
Like charges repel (push apart); unlike charges attract (pull together).
Objects get charged when electrons (tiny negative particles) move from one object to another.
[Diagram: phys-field-lines] - Available in full study mode
Three ways to charge something: Friction — rub two materials so electrons rub off one onto the other.
Contact — touch a charged object so charge flows across.
Induction — bring a charge near (no touching) and ground the object; it keeps the opposite charge.
| Method | What you do | What happens |
|---|---|---|
| Friction (rubbing) | Rub two different materials together (e.g. a cloth on a rod). | Electrons are transferred from one to the other — one ends up negative, the other positive. |
| Contact | Touch a charged object onto a neutral one. | Some charge flows across until they share it — both end up with the same sign of charge. |
| Induction | Bring a charge near (not touching), then ground the far side and remove the ground. | The object keeps an opposite charge to the one brought near — without ever touching it. |
Conservation of charge: Charge is never created or destroyed — it only moves.
If one object gains a charge of −q, the object it took electrons from is left with +q. The total stays the same.
Coulomb's law gives the force between two point charges. The force grows with the charges and shrinks with the square of the distance between them:
- electric force between the charges (N)
- Coulomb constant, 8.99 × 10⁹ N m² C⁻²
- the two point charges (C, coulombs)
- distance between the charges (m)
Which way does the force point?: The size comes from the formula; the direction comes from the signs.
Same signs (+ and +, or − and −) → the charges repel (push apart).
Opposite signs (+ and −) → the charges attract (pull together).
Worked example — force between two charges
Two point charges, q1 = 2.0 × 10⁻⁶ C and q2 = 3.0 × 10⁻⁶ C, sit 0.20 m apart. Find the force between them. (k = 8.99 × 10⁹ N m² C⁻².)
Solution
- Start with the given formula:
- Put in the numbers (q1 = 2.0 × 10⁻⁶, q2 = 3.0 × 10⁻⁶, r = 0.20):
- Work it out — keep the unit:
Final answer
F = 1.3 N. Both charges are positive, so it is a repulsive force (they push apart).
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How this is tested: Coulomb's law and charging open the Fields theme.
- Paper 1A: quick inverse-square reasoning — how the force changes when a charge is halved/doubled or the separation changes; or the sign of an induced charge after grounding. - Paper 2: calculate the force F = k q1q2/r² between point charges.
Classic trap: forgetting the square — halving the separation multiplies the force by 2² = 4, not by 2.
The scaling shortcut: Because F is proportional to q_{1}q_{2} and to 1/r², you don't always need k.
Halve one charge → F is halved. Double the separation → F is divided by 2² = 4. Combine the effects step by step.
IB-style question — (a) halving one charge
Two point charges a fixed distance apart exert a force of 8.0 N on each other. One of the charges is replaced by another of half its size, with the separation unchanged. Find the new force.
Solution
- Start with the given formula — F is proportional to each charge:
- Halving one charge halves the top of the fraction, so it halves F:
- The separation did not change, so nothing else affects F.
Final answer
Fnew = 4.0 N — half the original, because F is proportional to each charge.
IB-style question — (b) doubling the separation
Starting again from the original force of 8.0 N (both charges unchanged), the two charges are now moved to twice their original separation. Find the new force.
Solution
- Same given formula — F depends on 1/r²:
- Doubling the separation divides F by 2² = 4:
- Work it out — keep the unit:
Final answer
Fnew = 2.0 N — a quarter of the original, because F is proportional to 1/r².