Electric field and the field due to a point charge
Electric field at a point is the force per unit positive test charge there. For a point charge, E = q/(4πε₀r²), pointing radially away from a positive charge.
What is an electric field?
A charge changes the space around it. Another charge placed there feels a force. We say the first charge sets up an electric field. The electric field at a point is the force on a unit positive test charge placed there: E = F/q₀. The test charge q₀ must be very small, so that it does not disturb the charges that make the field. E is a vector. Its value depends on the source charges and on the point, not on the test charge. Once E is known, the force on any charge q at that point is simply F = qE.
Derivation: field due to a point charge
- 1Place a point charge q at the origin O. Choose a point P at distance r from O.
- 2Put a small positive test charge q₀ at P.
- 3By Coulomb's law, the force on q₀ is F = (1/4πε₀)(q q₀/r²) r̂, where r̂ is the unit vector from O to P.
- 4Field = force per unit test charge: E = F/q₀.
- 5★So, E = (1/4πε₀)(q/r²) r̂. It points radially outward for q > 0 and inward for q < 0. Its size is the same at all points on a sphere centred on q.
★ key step
Field due to a system of charges
- Find the field of each charge at the point, one at a time, using E = kq/r².
- Draw each field: away from a positive charge, towards a negative charge.
- Add the fields as vectors: E = E₁ + E₂ + ... + Eₙ.
- At the midpoint of two equal like charges, the fields cancel, so E = 0.
- At the midpoint of two equal unlike charges, the fields add, pointing towards the negative charge.
Worked example: field and force
Find the electric field at a point 20 cm from a charge of +8 nC in air. What force would an electron feel there? (e = 1.6 × 10⁻¹⁹ C)
- Given: q = 8 × 10⁻⁹ C, r = 0.20 m.
- E = kq/r² = 9 × 10⁹ × 8 × 10⁻⁹ ÷ (0.20)² = 72 ÷ 0.04 = 1800 N C⁻¹.
- Direction: away from the positive charge.
- Force on the electron: F = eE = 1.6 × 10⁻¹⁹ × 1800 = 2.88 × 10⁻¹⁶ N.
- The electron is negative, so the force points towards the charge.
E = 1.8 × 10³ N C⁻¹, away from the charge; force on the electron ≈ 2.9 × 10⁻¹⁶ N, towards the charge
Worked example: field at a midpoint
Charges of +3 μC and −3 μC are at A and B, 20 cm apart in air. Find the electric field at M, the midpoint of AB.
- Each charge is 0.10 m from M.
- Field of +3 μC at M: E₁ = 9 × 10⁹ × 3 × 10⁻⁶ ÷ (0.10)² = 2.7 × 10⁶ N C⁻¹, away from A, that is towards B.
- Field of −3 μC at M: E₂ = 2.7 × 10⁶ N C⁻¹, towards B (towards the negative charge).
- Both point the same way, so they add: E = 2 × 2.7 × 10⁶ = 5.4 × 10⁶ N C⁻¹.
5.4 × 10⁶ N C⁻¹, pointing from A towards B
How to draw the E versus r graph for a point charge
- Draw r on the horizontal axis and E on the vertical axis.
- Draw a curve that starts very high near r = 0 and falls steeply.
- Show it coming closer and closer to the r-axis without touching it.
- Label the curve E ∝ 1/r²: doubling r makes E one quarter.
Before lightning, the field between a thundercloud and the ground becomes huge. When it reaches about 3 × 10⁶ N C⁻¹, air stops being an insulator and a spark jumps across. Lightning tends to strike tall, lone objects, so never shelter under a lone tree in a monsoon storm.
1 marks: “Define electric field at a point. Give its SI unit.” The force per unit positive test charge placed at that point, E = F/q₀. Unit N C⁻¹ (or V m⁻¹).
2 marks: “An electron and a proton are placed in the same uniform electric field. Compare the forces on them and their accelerations.” The forces are equal in size (eE) and opposite in direction. Acceleration a = eE/m; the electron is about 1836 times lighter, so its acceleration is about 1836 times larger.
3 marks: “Charges +16 μC and +4 μC are 30 cm apart. Find the point on the line joining them where the electric field is zero.” Between them, x from the 16 μC charge: 16/x² = 4/(0.30 − x)². So 4/x = 2/(0.30 − x), giving x = 0.20 m. The point is 20 cm from the 16 μC charge.
E = F/q, and for one charge E = kq/r². The field points out of plus, into minus.