D. Gravitational, Electric, and Magnetic Fields · Gravitational, Electric and Magnetic FieldsElectric Fields of Point Charges and Field Lines
D2.1 D2.3 D3.3 D3.2About 75 minutes
Learning goals- I can define electric field strength as E = F/q and use E = kq/r² for a point charge.
- I can find the net electric field at a point from two or more charges, using vectors.
- I can draw and interpret field-line diagrams for single charges, pairs of charges and charged conductors.
- I can explain why the field inside a conductor is zero and why charge gathers at sharp points.
Words to know
electric field
The region around a charged object in which another charge feels an electric force.
electric field strength (E)
The electric force per unit positive charge at a point: E = F/q, in N/C. Its direction is the direction of the force on a positive test charge.
test charge
A small positive charge used (in imagination) to probe a field without disturbing it.
electric dipole
A pair of equal and opposite charges separated by a small distance.
electrostatic equilibrium
The state of a conductor in which no charges are moving. Then the field inside the conductor is zero and any excess charge is on its surface.
Faraday cage
A conducting enclosure that shields its inside from external electric fields.
From force to field
In Gravitational Fields you replaced "Earth pulls on the Moon" with "Earth makes a field, and the field pulls on the Moon". Electric forces work the same way, and here the field idea is even more useful, because charges can be positive or negative and can move.
Michael Faraday, who invented field lines in the 1830s and 1840s, had little formal mathematics. He thought in pictures: lines of force filling space around charges and magnets. James Clerk Maxwell later turned Faraday's pictures into equations, and those equations predicted light (you will see how in The Wave Nature of Light).
Electric field strength
Place a small positive test charge q at a point. If it feels an electric force F, the field strength there is
E = F/q (units N/C)
E is a vector pointing in the direction of the force on a positive charge. Turned around, the force on any charge q in a field E is
F = qE
A negative charge feels a force opposite to the field direction.
For a point charge Q, Coulomb's law gives F = kQq/r², so
E = kQ/r²
- The field points away from a positive Q and toward a negative Q.
- Like g = GM/r², it depends only on the source charge and the distance, not on the test charge.
- It follows an inverse-square law: double the distance, one quarter the field.
Worked exampleExample 1: Field of a point charge
Find the electric field 0.30 m from a +5.0 μC point charge, and the force on an electron placed there.
- E = kQ/r² = (8.99 × 10⁹)(5.0 × 10⁻⁶)/(0.30)² = 4.495 × 10⁴/0.090 = 4.99 × 10⁵ N/C.
- Direction: away from the positive charge.
- F = qE = (1.60 × 10⁻¹⁹)(4.99 × 10⁵) = 7.99 × 10⁻¹⁴ N.
- The electron is negative, so the force is opposite to E: toward the +5.0 μC charge.
E ≈ 5.0 × 10⁵ N/C, away from the charge; F ≈ 8.0 × 10⁻¹⁴ N toward the charge. (That tiny force gives the electron an acceleration of about 9 × 10¹⁶ m/s², because its mass is so small.)
Adding fields: superposition
The field at a point from several charges is the vector sum of the fields from each charge alone. The method:
- Draw the point and each charge.
- For each charge, find the size E = k|Q|/r² and draw its direction (away from +, toward −).
- Add the vectors (components, or Pythagoras when they are perpendicular).
Worked exampleExample 2: Field on the axis of a dipole
Charges of +4.0 μC and −4.0 μC are 0.20 m apart. Find the field at the midpoint.
- Each charge is 0.10 m away. Each field has size E = (8.99 × 10⁹)(4.0 × 10⁻⁶)/(0.10)² = 3.60 × 10⁶ N/C.
- The field from the + charge points away from it, toward the − charge. The field from the − charge points toward it, also toward the − charge.
- Both point the same way, so add: 2 × 3.60 × 10⁶ = 7.19 × 10⁶ N/C.
7.2 × 10⁶ N/C, pointing toward the negative charge.
Worked exampleExample 3: Field off the axis
A +3.0 nC charge is at (0, 0) and a −3.0 nC charge is at (0.40 m, 0). Find the field at P (0.20 m, 0.15 m).
- Each charge is √(0.20² + 0.15²) = 0.25 m from P.
- Size of each field: E = (8.99 × 10⁹)(3.0 × 10⁻⁹)/(0.25)² = 432 N/C.
- E from the + charge points away from it, along (0.20, 0.15)/0.25 = (0.8, 0.6): components (+345, +259) N/C.
- E from the − charge points toward it, from P to (0.40, 0): direction (0.20, −0.15)/0.25 = (0.8, −0.6): components (+345, −259) N/C.
- Add: E_x = 690 N/C, E_y = 0.
690 N/C in the +x direction, parallel to the line from the + charge to the − charge. The vertical parts cancel by symmetry.
Rules for electric field lines
- Lines start on positive charges and end on negative charges (or go off to infinity).
- The number of lines leaving or entering a charge is proportional to the size of the charge.
- Lines never cross.
- The field is strongest where the lines are closest together.
- The field at any point is tangent to the line through it.
- Lines meet a conductor's surface at right angles (when charges are not moving).
Compare with gravity: gravitational field lines only end on masses, because there is no negative mass. Electric field lines both start and end on charges.
Fields and conductors
In a conductor, electrons move freely. Put a conductor in a field and the electrons shift until the field they create cancels the external field inside the metal. Then nothing moves: the conductor is in electrostatic equilibrium. At that point:
- E = 0 everywhere inside the conductor material. (If it weren't zero, electrons would still be moving.)
- Any excess charge sits on the outer surface, as far apart as it can get.
- The field just outside is perpendicular to the surface.
- On an irregular conductor, charge crowds at sharp points, so the field is strongest there. Lightning rods and the sharp points on the high-voltage equipment in power stations use or avoid this effect.
A hollow conducting shell protects its inside from outside electric fields. This Faraday cage effect is why a car with a metal body offers protection in a lightning storm, why a phone may lose signal in an elevator, and why a microwave oven has a metal mesh in its door.
Just for funIn the 1740s Benjamin Franklin had to choose which kind of charge to call "positive". He picked the one we now know is a shortage of electrons, so electrons ended up negative. Physics students have been drawing current arrows the "wrong" way ever since.
Hands-on activityMap field shapes with a simulation, and see shielding at home
Part A. Open the free PhET simulation "Charges and Fields" (phet.colorado.edu).
- Place one +1 nC charge. Turn on "Electric Field" and use the E-field sensor at 0.5 m, 1.0 m and 1.5 m. Check that E drops to 1/4 and 1/9.
- Build a dipole (+1 nC and −1 nC). Sketch the field arrows. Then build two +1 nC charges and find the point where the sensor reads zero.
- Try +2 nC and −1 nC. Where is the field zero now? (Hint: outside the pair, beyond the smaller charge.)
Part B. You need a battery-powered AM/FM radio (or a phone with an FM radio feature) and a large sheet of aluminium foil, or a metal cookie tin with a lid.
- Tune the radio to a station and turn it up a little.
- Wrap it completely in foil (or close it in the tin). Listen.
- Leave a small gap in the foil and listen again.
What to look for: Part A: with +2 nC and −1 nC 1 m apart, the zero point is on the far side of the −1 nC charge, about 2.4 m beyond it (3.4 m from the +2 nC charge). Part B: the station fades or disappears when the radio is fully enclosed in metal, because the conductor shields the inside from the radio waves' changing electric fields. A gap lets some signal back in. (Do not put a phone in a microwave oven or turn the oven on to test this.)
Your turn
1Define electric field strength and give its unit.
The electric force per unit positive test charge at a point, E = F/q, in N/C.
2Find the field 0.50 m from a −2.0 μC charge.
E = (8.99 × 10⁹)(2.0 × 10⁻⁶)/(0.50)² = 7.2 × 10⁴ N/C, directed toward the charge.
3A 2.0 nC charge feels a force of 6.0 × 10⁻⁵ N [E]. Find the field. What force would a −4.0 nC charge feel at the same point?
E = F/q = 6.0 × 10⁻⁵/2.0 × 10⁻⁹ = 3.0 × 10⁴ N/C [E]. F = qE = (4.0 × 10⁻⁹)(3.0 × 10⁴) = 1.2 × 10⁻⁴ N [W] (opposite, since the charge is negative).
4A +2.0 μC charge is at x = 0 and a −3.0 μC charge is at x = 0.50 m. Find the field at x = 0.20 m.
From +2.0 μC (0.20 m): 4.50 × 10⁵ N/C, pointing +x (away). From −3.0 μC (0.30 m): 3.00 × 10⁵ N/C, pointing +x (toward it). Total 7.5 × 10⁵ N/C in the +x direction.
5What charge produces a field of 1.0 × 10⁵ N/C at a distance of 0.50 m?
Q = Er²/k = (1.0 × 10⁵)(0.25)/(8.99 × 10⁹) = 2.8 × 10⁻⁶ C (2.8 μC).
6Sketch the field lines around two equal negative charges. Where is the field zero?
Lines come in from far away and end on each charge; they bend away from the region between the charges. The field is zero at the midpoint. (The picture is the same as for two positive charges, with every arrow reversed.)
More practice: warm-up
1Which way does the field point near a positive charge?
2A negative charge is placed in a field pointing north. Which way is the force on it?
3Where do electric field lines begin and end?
They begin on positive charges and end on negative charges (or at infinity).
4What is the electric field inside a conductor in electrostatic equilibrium?
5If the distance from a point charge doubles, the field becomes ___ as large.
More practice: on your own
1Charges of +2.0 nC at (0, 0.30 m) and +4.0 nC at (0.40 m, 0). Find the field at the origin.
From the +2.0 nC charge (0.30 m above): 200 N/C pointing down (−y). From the +4.0 nC charge (0.40 m to the right): 225 N/C pointing left (−x). E = √(200² + 225²) = 301 N/C, at tan⁻¹(200/225) = 42° below the −x axis (pointing left and down).
2A +4.0 μC charge is at x = 0 and a −1.0 μC charge at x = 0.30 m. Where on the x-axis is the field zero?
It must be outside the pair, beyond the smaller charge (x > 0.30 m), where the two fields point opposite ways and can balance: 4/x² = 1/(x − 0.30)², so 2(x − 0.30) = x, x = 0.60 m.
3How strong must an upward field be to hold an electron at rest against gravity? A proton?
eE = mg. Electron: E = (9.11 × 10⁻³¹)(9.8)/(1.60 × 10⁻¹⁹) = 5.6 × 10⁻¹¹ N/C, pointing down (the force on an electron is opposite to E, so E must point down for an upward force). Proton: E = (1.67 × 10⁻²⁷)(9.8)/(1.60 × 10⁻¹⁹) = 1.0 × 10⁻⁷ N/C, pointing up. Gravity is negligible for particles in almost any real field.
4Why do field lines never cross?
At a crossing point the field would have two directions at once, which is impossible: a test charge there could only feel one net force.
5Explain why a car with a metal body is a relatively safe place in a lightning storm.
The metal body acts as a Faraday cage: charge moves along the outer metal surface and the field inside stays close to zero, so the current does not pass through the people inside. (Stay inside and do not touch metal parts; a convertible or fibreglass car does not give this protection.)
Challenge
1Air breaks down (sparks) when the field reaches about 3 × 10⁶ N/C. What is the maximum charge a metal sphere of radius 0.15 m can hold in air? (Outside a sphere, E = kQ/r².)
At the surface, E = kQ/R², so Q = ER²/k = (3 × 10⁶)(0.15)²/(8.99 × 10⁹) = 7.5 × 10⁻⁶ C (about 7.5 μC).
2Four equal positive charges sit at the corners of a square. What is the field at the centre? What if one of them were removed?
With all four, the fields from opposite corners cancel in pairs: E = 0. Remove one: its partner's field is no longer cancelled, so E = kQ/r² (r = half the diagonal), pointing toward the empty corner (away from the charge opposite it).
3A charge +Q is at the origin and −Q is at x = d. Show that far along the perpendicular bisector the field points parallel to the line from +Q to −Q, and explain why it weakens faster than 1/r².
By symmetry (as in Example 3), the components perpendicular to the axis cancel and the components along the axis add, pointing from +Q toward −Q. Far away, the two charges look almost like they are at the same place, so their fields nearly cancel; only the small difference in their directions survives, and that difference shrinks as d/r. So the field falls off faster than 1/r² (it actually goes as 1/r³ for a dipole).
Quiz
1. The direction of the electric field at a point is defined as
2. The field 2.0 m from a charge is 400 N/C. At 4.0 m it is
3. Which statement about electric field lines is false?
4. A proton and an electron are placed at the same point in an electric field. Compared with the force on the proton, the force on the electron is
5. Inside a hollow metal box with charge on it, in electrostatic equilibrium, the electric field is
6. Two equal positive charges are 10 cm apart. The electric field is zero
Big idea (D2.3, D3.3): an electric field E = F/q describes the space around charges. For a point charge E = kQ/r². Field lines show direction (away from +, toward −) and strength (spacing). Conductors have zero field inside and gather charge at sharp points.
How to help: Ask your teen to sketch field lines for one charge, a dipole and two like charges from memory, and to explain the rules they used. The radio-in-foil test is a quick, safe demonstration of shielding.
Common difficulties:
- Forgetting that the force on a negative charge is opposite to E.
- Adding the sizes of fields that point in different directions.
- Drawing lines that cross, or lines that start on negative charges.
- Mixing up E (field, N/C) with F (force, N): F = qE.
Safety: never put electronics in a microwave oven. Foil and tins are safe to use with a battery radio.