Electric field
What is an electric field?
A charge Q sits at the origin. Put a small test charge q at a point P, a distance r away. By Coulomb’s law it feels a force F equals k Q q over r², along r̂.
The force depends on q. Double q and the force doubles. So divide by q. What is left, k Q over r², depends only on Q and on where P is.
This is the electric field of Q at P: E equals F over q, which is k Q over r² r̂. It is the force a unit positive charge would feel there. Its unit is newton per coulomb.
The test charge must be very small, so that it does not push Q out of place. Strictly, E is the limit of F over q as q goes to zero.
If Q is positive, E points straight out from Q at every point. If Q is negative, it points straight in. On a sphere centred on Q, E has the same size everywhere: the field is spherically symmetric.
Once you know E at a point, you know the force on any charge placed there: F equals q E. A positive charge is pushed along E, a negative charge the opposite way.
For many charges, use superposition again. The field at P is the vector sum of the fields of each charge: E equals E₁ plus E₂ plus and so on.
Is the field just a clever trick? No. If a charge suddenly moves, a distant charge does not feel the change at once. The news travels outwards through the field, at the speed of light.
So the field is something physical: it carries the effect from one charge to another. Changing electric and magnetic fields even travel as waves, as you will see in a later chapter.
Fields in numbers
Two point charges: q₁ equals plus 10 nanocoulombs and q₂ equals minus 10 nanocoulombs, 10 centimetres apart. Find the field at three points.
Point A is midway, 5 centimetres from each. Field of q₁: 9 × 10⁹ times 10⁻⁸, over 0.05 squared. That is 90 over 0.0025: 3.6 × 10⁴ newtons per coulomb, pointing away from q₁, towards q₂.
Field of q₂ at A has the same size and points towards q₂, the same way. So they add: 7.2 × 10⁴ newtons per coulomb, from q₁ towards q₂.
Point B is on the line, 5 centimetres beyond q₁, on the far side from q₂. q₁ gives 3.6 × 10⁴, pointing away from q₁. q₂ is 15 centimetres away: 90 over 0.0225, which is 4 × 10³, pointing back towards q₂.
They are opposite, so subtract: 3.6 × 10⁴ minus 0.4 × 10⁴ gives 3.2 × 10⁴ newtons per coulomb, pointing away from q₂.
Point C is 10 centimetres from both charges. Each field is 90 over 0.01, which is 9 × 10³. They are 120° apart, so the resultant is also 9 × 10³ newtons per coulomb, parallel to the line from q₁ to q₂.
Now a charge moving in a field. An electron falls 1.5 centimetres in a uniform field of 2.0 × 10⁴ newtons per coulomb. Its acceleration is e E over m: 1.6 × 10⁻¹⁹ times 2 × 10⁴, over 9.1 × 10⁻³¹, about 3.5 × 10¹⁵ metres per second squared.
From h equals ½ a t², the time is the square root of 2h over a: about 2.9 × 10⁻⁹ seconds. A proton in the same field, reversed, has the same force but about 1840 times the mass. It takes about 1.3 × 10⁻⁷ seconds.
Gravity’s 9.8 metres per second squared is nothing beside these accelerations, so we can ignore it.
Try it: the field around charges
- P starts at point A of the example. Each charge gives 3.6 × 10⁴ N/C, and the net field is 7.2 × 10⁴ N/C.
- Drag P to x = −10 cm, 5 cm beyond q₁ (point B). The net field drops to 3.2 × 10⁴ N/C, pointing away from q₂.
- Make q₂ positive (+10 nC). Where is the field zero now?
- The small blue arrows show the direction of E everywhere; brighter arrows mean a stronger field.
The third point
Charges +10 nC and −10 nC are 10 cm apart. What is the field at a point 10 cm from each charge?
- 9 × 10³ N/C, parallel to the line from the + charge to the − charge
- 1.8 × 10⁴ N/C, parallel to the line joining them
- 1.56 × 10⁴ N/C, along the perpendicular bisector
- Zero, because the charges are equal and opposite
Answer
A. Each field has size 9 × 10⁹ × 10⁻⁸ / (0.1)² = 9 × 10³ N/C: away from +q and towards −q. The point makes an equilateral triangle with the charges, so the two fields are 120° apart. Two equal vectors at 120° add to one of the same size: 9 × 10³ N/C, parallel to the line from +q to −q. (1.8 × 10⁴ adds the sizes; 1.56 × 10⁴ uses 60°.)
Force on an electron
At a point, the electric field is 2 × 10⁴ N/C pointing east. What force acts on an electron placed there?
- 3.2 × 10⁻¹⁵ N, east
- 3.2 × 10⁻¹⁵ N, west
- 1.25 × 10²³ N, west
- No force: the field belongs to other charges
Answer
B. F = qE = 1.6 × 10⁻¹⁹ × 2 × 10⁴ = 3.2 × 10⁻¹⁵ N. The electron is negative, so the force is opposite to E: west. (1.25 × 10²³ comes from dividing E by e instead of multiplying.)
Summary
- Definition (small test charge q):
E = F / q - Point charge (along r̂):
E = kQ / r² - Force on a charge:
F = qE - Superposition:
E = E₁ + E₂ + … + Eₙ
- The electric field at a point is the force per unit positive charge there; unit N/C.
- The field of a point charge points away from +Q and towards −Q, and falls as 1/r².
- The field of several charges is the vector sum of their separate fields.
- Fields carry the effect of one charge to another, at a finite speed; they are physical, not just a trick.