- Explain electrification, electrostatic induction and the electroscope; apply q = N · e and conservation of charge (including sharing by contact)
- Calculate forces with Coulomb’s law and analyse how they change
- Find the strength and direction of the field of a point charge and read field-line pictures
- Use superposition to find the resulting field and the point where it is zero
Rub a balloon on your hair and it sticks to the wall; take off a wool jumper in winter and tiny sparks crackle; lightning is a giant spark between a cloud and the ground. All of this is caused by electric charges and their field. The branch of physics that studies charges at rest is called electrostatics. In this lesson we study the properties of charge, the interaction of charges (Coulomb’s law) and the electric field — the topics of the DİM chapter «Electric charge. Electric field».
Electric charge and its properties
There are two kinds of charge: positive and negative. Like charges repel, unlike charges attract. In an atom the positive charge is on the protons and the negative charge on the electrons. Electrification by rubbing does not create charge — electrons move from one body to the other, so both bodies get charges equal in size and opposite in sign: glass rubbed with silk becomes positive, ebonite rubbed with wool negative. Any charge is a whole multiple of the elementary charge.
- qcharge of the body, C
- Nnumber of extra (or missing) electrons — a whole number
- eelementary charge, 1.6 · 10⁻¹⁹ C
Extra electrons mean a negative charge, missing electrons a positive one. Conservation of charge: in a closed system the algebraic sum of charges does not change: q₁ + q₂ + … = const.
1) A body has a charge of −3.2 · 10⁻¹⁸ C. How many extra electrons does it have?
2) A glass rod rubbed with silk lost 5 · 10¹⁰ electrons. What are the charges of the rod and the silk?
3) Can a body have a charge of 2.4 · 10⁻¹⁹ C?
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2) q = N · e = 5 · 10¹⁰ · 1.6 · 10⁻¹⁹ = 8 · 10⁻⁹ C: the rod has +8 nC, the silk −8 nC (charge is conserved).
3) 2.4 · 10⁻¹⁹ / (1.6 · 10⁻¹⁹) = 1.5 — not a whole number, so it cannot.
When two identical metal balls touch, the charge is shared equally (the balls are the same, neither is preferred), while the total is conserved.
- q₁, q₂charges before contact (with their signs)
- q₁′, q₂′charges after contact
Only for conducting balls of the same size and shape. Mind the signs: +8 and −2 → (8 − 2) / 2 = +3.
1) Two identical balls carry +8 nC and −2 nC. What is the charge of each after they touch?
2) Identical balls with +6 nC and −6 nC are brought into contact. Result?
3) Ball A with charge q touches the uncharged ball B and then the uncharged ball C (all identical). What is A’s final charge?
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2) (6 − 6) / 2 = 0 — both become neutral.
3) A with B: q / 2 each. A with C: (q / 2 + 0) / 2 = q / 4.
Bring a charged body close to a conductor without touching it. The free electrons in the conductor redistribute: the near side gets a charge opposite to the body’s, the far side a charge of the same sign, while the conductor as a whole stays neutral. This is electrostatic induction; the charges arrange themselves so that the field inside the conductor is zero. If you touch the conductor (earth it) and remove your hand before taking the charged body away, the conductor keeps a charge opposite to the body’s. A dielectric has no free electrons, but its molecules polarise — that is why a charged comb attracts bits of paper.
A device that shows whether a body is charged: a metal rod with two light leaves at its lower end. When the rod is charged, the leaves get charges of the same sign and spread apart; the wider they spread, the larger the charge. An electroscope with a scale is called an electrometer.
1) A positively charged rod is brought close to the ball of a negatively charged electroscope without touching it. What happens to the leaves?
A) they spread wider B) they come closer together C) nothing D) the leaves attract each other E) the electroscope discharges
2) What would be seen if the rod were negative?
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2) A negative rod pushes electrons down into the leaves — they spread wider. This is how the sign of an unknown charge is found.
Coulomb’s law
In 1785 Charles Coulomb measured the interaction of point charges with a torsion balance. He found that the force is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them, and that it acts along the line joining them. In a medium (for example, water) the force is ε times weaker than in vacuum.
- Fforce of interaction, N
- k9 · 10⁹ N · m²/C² (k = 1 / (4π · ε₀))
- εrelative permittivity of the medium (1 for vacuum and air, 81 for water)
- rdistance between the charges, m
For point charges (bodies much smaller than the distance between them). By Newton’s third law F₁₂ = F₂₁ — even when the charges differ, they act on each other with the same force.
1) Charges of 2 µC and 3 µC are 30 cm apart in air. Find the force between them.
2) How does the force change if a) the distance is halved; b) both charges and the distance are doubled; c) the charges are moved from air into water (ε = 81)?
3) Two charges of 1 µC each repel with a force of 0.1 N in air. How far apart are they?
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F = 9 · 10⁹ · 2 · 10⁻⁶ · 3 · 10⁻⁶ / 0.3² = 0.054 / 0.09 = 0.6 N (like charges — repulsion).
2) a) r² drops 4 times → F increases 4 times; b) the numerator grows 2 · 2 = 4 times and the denominator 2² = 4 times → no change; c) 81 times smaller.
3) r = √(k · q₁ · q₂ / F) = √(9 · 10⁹ · 10⁻¹² / 0.1) = √0.09 = 0.3 m.
Point charges q₁ = q and q₂ = 4q are a distance r apart. What is the ratio F₂ / F₁ of the force on q₂ to the force on q₁?
A) 4 B) 1/4 C) 1 D) 2 E) 16
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Option A is the mistake “the bigger charge is pushed harder”.
Two identical metal balls carry +6 nC and −2 nC; at a distance r they attract with force F₁. They are brought into contact and returned to the same distance; now the force is F₂. Calculate the ratio F₁ / F₂.
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F₁ ~ 6 · 2 = 12, F₂ ~ 2 · 2 = 4 (k and r are the same).
F₁ / F₂ = 12 / 4 = 3. Answer: 3.
Note: the direction changed too — attraction before, repulsion now.
The electric field and its strength
Charges do not act on each other “directly” across empty space: each charge creates an electric field around itself, and the field exerts forces on other charges. To study a field we bring a small positive test charge q₀ into it. The force on the test charge is proportional to q₀, so the ratio F / q₀ depends only on the field — this is the field strength (intensity).
- Efield strength, N/C (= V/m)
- q₀test charge placed in the field, C
- qpoint charge creating the field, C
The field strength is a vector: it points away from a positive charge and towards a negative one. A charge q in the field feels F = q · E; the force on a negative charge (an electron) points opposite to E.
1) Find the field strength in air 20 cm from a 4 nC point charge. What force acts on an electron placed there?
2) How does the field strength change if the distance to the charge is tripled?
3) An oil drop of mass 4.8 · 10⁻¹⁵ kg hangs motionless in a uniform downward field of 10⁵ N/C. What is its charge, and how many extra electrons does it carry? (g = 10 m/s²)
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2) E ~ 1/r² → 9 times smaller.
3) Balance: |q| · E = m · g → |q| = 4.8 · 10⁻¹⁵ · 10 / 10⁵ = 4.8 · 10⁻¹⁹ C. The field points down and the force up, so the drop is negative: N = 4.8 · 10⁻¹⁹ / (1.6 · 10⁻¹⁹) = 3 electrons. This is the idea of Millikan’s experiment: the charge always comes out as a whole multiple of e.
A field is drawn with field lines: the tangent to a line at any point shows the direction of E there.
- Lines start on positive charges and end on negative ones (or at infinity) — electrostatic field lines are not closed.
- Lines never cross: at each point the field has only one direction.
- Where the lines are dense, the field is strong; they crowd together near a point charge.
- In a uniform field (for example, between the plates of a parallel-plate capacitor) the lines are parallel and evenly spaced.
The superposition principle
In the field of several charges each charge makes its own field independently of the others, and the resulting field is their vector sum. First find the field of each charge on its own (magnitude k · |q| / r², direction by the sign of the charge), then add the vectors.
- E⃗₁, E⃗₂field strengths created by each charge alone
Along one line: vectors in the same direction add, opposite ones subtract. When they are perpendicular, E = √(E₁² + E₂²).
1) Charges +2 nC and −2 nC are 6 cm apart in air. Find the field strength at the midpoint of the segment joining them.
2) A right triangle has legs of 3 cm; charges +3 nC and +4 nC sit at the acute-angle vertices. What is the field strength at the right-angle vertex?
3) What is the field strength midway between two equal charges +q?
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2) E₁ = 9 · 10⁹ · 3 · 10⁻⁹ / 0.03² = 3 · 10⁴ N/C, E₂ = 9 · 10⁹ · 4 · 10⁻⁹ / 0.03² = 4 · 10⁴ N/C, along the legs, perpendicular: E = √(3² + 4²) · 10⁴ = 5 · 10⁴ N/C.
3) The vectors are equal and opposite → E = 0.
On an axis, charge +q is at x = 0 and charge −4q at x = 3 cm. At which point is the field strength zero?
A) x = 1 cm B) x = 2 cm C) x = −3 cm D) x = 6 cm E) x = −1 cm
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2) The zero point must be outside the segment, on the side of the smaller charge (the larger charge’s field weakens with distance), so x < 0.
3) Let d be the distance from +q: k · q / d² = k · 4q / (d + 3)² → d + 3 = 2d → d = 3 cm, x = −3 cm (C).
Note: for like charges +q and +4q the zero point would lie between them, 1 cm from the smaller one (3 − d = 2d).
When the field moves a charge from one point to another it does work — this leads to the ideas of potential and voltage: see the lesson «Electric potential and potential difference». Devices that store charge and field energy are in the lesson «Capacitance and capacitors».
Key points
- Charge is quantised (q = N · e, e = 1.6 · 10⁻¹⁹ C) and conserved; identical balls in contact share it equally.
- In electrostatic induction a conductor’s charges redistribute; an electroscope detects charge and its sign.
- Coulomb’s law: F = k · |q₁| · |q₂| / (ε · r²), k = 9 · 10⁹ N · m²/C²; F₁₂ = F₂₁.
- Field strength E = F / q₀, for a point charge E = k · |q| / (ε · r²); unit N/C = V/m; it points away from positive and towards negative charges.
- Superposition: E⃗ = E⃗₁ + E⃗₂ + …; the point where E = 0 is between like charges and outside unlike ones, nearer the smaller charge.
Check yourself
12 questions. Every correct answer earns XP.