D. Gravitational, Electric, and Magnetic Fields · Gravitational, Electric and Magnetic FieldsUniform Electric Fields Between Parallel Plates
D2.3 D3.3 D3.2 D2.1About 70 minutes
Learning goals- I can describe the field between oppositely charged parallel plates and draw its field and equipotential lines.
- I can derive and use E = ΔV/d for a uniform field.
- I can find forces, potential differences and energy changes for charges between parallel plates.
- I can compare the uniform field between plates with the near-Earth gravitational field, and contrast it with the field of a point charge.
Words to know
parallel plates
Two flat conducting plates facing each other with equal and opposite charges. Between them the field is uniform.
uniform electric field
A field with the same strength and direction at every point, shown by parallel, evenly spaced field lines.
E = ΔV/d
The field strength between parallel plates: potential difference divided by plate separation. Units V/m, which equal N/C.
fringing
The bending and weakening of the field near the edges of parallel plates.
capacitor
A device made of two conductors separated by an insulator, used to store charge and energy. Parallel plates are the simplest kind.
Making a field you can control
The field of a point charge changes everywhere: it gets weaker with distance and points in a different direction at every point. That makes it awkward for experiments. Physicists and engineers usually want a field that is the same everywhere in a region. The simplest way to make one: two flat metal plates, connected to the terminals of a power supply.
Uniform fields are inside every capacitor in your phone, in touchscreens, in the deflection plates of older oscilloscope tubes, in inkjet printers, and in Millikan's famous experiment in the next lesson. This lesson builds the tools; Charged Particles in Uniform Electric Fields puts them to work.
The field between the plates
Connect two parallel plates to a battery. One plate gets positive charge and the other an equal negative charge. If the plates are large compared with the gap between them:
- Between the plates, the field lines are straight, parallel and evenly spaced, running from the + plate to the − plate. The field is uniform.
- Outside the plates, the fields from the two plates almost cancel, so E ≈ 0.
- Near the edges, the lines bow outward and the field weakens. This fringing is usually ignored when the plates are much wider than the gap.
The field strength does not depend on where you are between the plates: a charge feels the same force right next to a plate as in the middle. That is very different from a point charge.
E = ΔV/d
Move a positive test charge q from the − plate to the + plate, against the field, at constant speed. You must push with force F = qE through distance d, so the work is W = qEd. That work equals the gain in electric potential energy, qΔV:
qEd = qΔV → E = ΔV/d
So the field can be given in volts per metre, and 1 V/m = 1 N/C. (Check: 1 V/m = 1 J/(C·m) = 1 N·m/(C·m) = 1 N/C.)
Two consequences:
- Across the gap, the potential changes steadily (linearly) from one plate to the other. Halfway across, you have half the voltage.
- The equipotentials are flat planes parallel to the plates, evenly spaced.
Worked exampleExample 1: A simple pair of plates
Two parallel plates 2.0 cm apart are connected to a 120 V supply. (a) Find the field. (b) Find the force on an electron between the plates. (c) Find the potential 0.50 cm from the negative plate.
- (a) E = ΔV/d = 120 V/0.020 m = 6.0 × 10³ V/m (N/C), from the + plate toward the − plate.
- (b) F = qE = (1.60 × 10⁻¹⁹)(6.0 × 10³) = 9.6 × 10⁻¹⁶ N. The electron is negative, so the force points toward the + plate.
- (c) V rises steadily from 0 at the − plate: V = Ex = (6.0 × 10³)(0.0050) = 30 V.
(a) 6.0 × 10³ N/C; (b) 9.6 × 10⁻¹⁶ N toward the positive plate, the same anywhere between the plates; (c) 30 V.
Worked exampleExample 2: Energy across the gap
In Example 1, a proton is released from rest next to the + plate. How much kinetic energy does it have when it reaches the − plate, in joules and electron volts?
- The proton moves from V = 120 V to V = 0, so ΔV = −120 V.
- ΔEk = −qΔV = −(1.60 × 10⁻¹⁹)(−120) = 1.92 × 10⁻¹⁷ J.
- In eV: (1 elementary charge)(120 V) = 120 eV.
1.92 × 10⁻¹⁷ J = 120 eV. Notice that the plate separation did not matter for the energy: only the potential difference does.
Fields compared
The region between parallel plates is the electric version of the region just above the ground.
- Near-Earth gravity: uniform g ≈ 9.8 N/kg, force F = mg, energy change ΔEg = mgΔh.
- Between plates: uniform E = ΔV/d, force F = qE, energy change ΔEe = qEΔd = qΔV.
Both are the "zoomed-in" view of a bigger field. Earth's field is radial when seen from space but uniform in a room. A point charge's field is radial, but the field of a very large flat sheet of charge is uniform. (The field of a large sheet does not weaken with distance, because as you move away, you "see" more of the sheet.)
One key difference: gravity's uniform field always points down toward the mass, while the electric field direction depends on which plate is positive, and positive and negative charges are pushed opposite ways.
What sets the field? Charge, area and the battery
The field between plates depends on how densely the charge is packed on them: E = Q/(ε₀A), where Q is the charge on one plate, A is the plate area and ε₀ = 8.85 × 10⁻¹² C²/(N·m²).
- Plates connected to a battery: ΔV is fixed. Pull the plates apart and E = ΔV/d drops.
- Plates charged and disconnected: Q is fixed, so E = Q/(ε₀A) stays the same as you pull them apart, and ΔV = Ed grows.
This is the basis of a capacitor. Many phone touchscreens are capacitive: your finger, a conductor, changes the field and the stored charge at the spot you touch, and the phone's circuits detect the change.
Just for funWhy did the electron love the space between the parallel plates? Because no matter where it went, it felt the same pull. Very stable relationship.
Hands-on activityBuild a simple capacitor and explore a uniform field
You need: two sheets of aluminium foil (about 20 cm × 20 cm), one sheet of plastic wrap or a thin plastic sheet slightly larger than the foil, two heavy books, a balloon or plastic comb, a wool sweater, and a small piece of tissue paper. Optional: a PhET simulation, "Capacitor Lab: Basics".
- Lay one foil sheet on a book, cover it with the plastic, and put the second foil sheet on top. Weigh it down with the other book, leaving a corner of each foil sheet sticking out. You have built a parallel-plate capacitor.
- Charge a balloon on the sweater and rub it against the top foil's corner several times while touching the bottom foil's corner with your other hand (this connects it to you, a large conductor).
- Lift the top book off and bring a tiny scrap of tissue near the top foil. Then separate the plates by lifting the top foil by its plastic edge (you may hear or feel a tiny spark).
- In the simulation, set a voltage, change the plate separation and watch the field and the stored charge.
What to look for: the top foil attracts the tissue, showing it holds charge. In the simulation, with the battery connected, increasing d lowers E (E = ΔV/d) and the stored charge drops. Disconnect the battery and increase d: the charge stays the same, E stays the same, and the voltage rises. Note: the static sparks from a balloon are harmless, but never connect home-made foil plates to a wall outlet or to anything except a small battery.
Your turn
1Two plates 5.0 mm apart have a field of 2.0 × 10⁴ N/C between them. What is the potential difference?
ΔV = Ed = (2.0 × 10⁴)(5.0 × 10⁻³) = 100 V.
1 V/m = 1 (J/C)/m = 1 (N·m)/(C·m) = 1 N/C.
3Plates 1.5 cm apart are connected to a 300 V supply. Find the force on a +3.0 nC charge between them.
E = 300/0.015 = 2.0 × 10⁴ N/C. F = qE = (3.0 × 10⁻⁹)(2.0 × 10⁴) = 6.0 × 10⁻⁵ N, toward the negative plate.
4In the previous question, how much work is needed to move the charge from the negative plate to the positive plate?
W = qΔV = (3.0 × 10⁻⁹)(300) = 9.0 × 10⁻⁷ J (equivalently F × d = 6.0 × 10⁻⁵ × 0.015).
5An electron starts from rest at the negative plate of a 50 V pair of plates. What kinetic energy does it have at the positive plate, in eV and J?
50 eV = 50 × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹⁸ J.
6Sketch the field and equipotentials between two parallel plates and explain how the drawing shows the field is uniform.
Straight, parallel, evenly spaced field lines from the + plate to the − plate, and evenly spaced equipotential lines parallel to the plates. Equal spacing means the same strength everywhere; parallel lines mean the same direction everywhere.
More practice: warm-up
1What shape are the field lines between parallel plates?
Straight, parallel and evenly spaced, from + to −.
2Write the formula for the field between parallel plates.
3What is the field like just outside the plates (away from the edges)?
4Does a charge feel a stronger force near one plate than in the middle?
No. The field (and force) is the same everywhere between the plates.
5Halfway between plates at 0 V and 80 V, what is the potential?
More practice: on your own
1Plates connected to a 200 V battery are 4.0 mm apart. They are moved to 8.0 mm apart, still connected. What happens to E?
E = ΔV/d: d doubles with ΔV fixed, so E halves, from 5.0 × 10⁴ to 2.5 × 10⁴ N/C.
2The same plates are charged, disconnected from the battery and then moved from 4.0 mm to 8.0 mm apart. What happens to E and ΔV?
The charge is fixed, so E = Q/(ε₀A) stays 5.0 × 10⁴ N/C. ΔV = Ed doubles to 400 V.
3Each plate has an area of 0.010 m² and carries 1.0 × 10⁻⁸ C. Find E between them.
E = Q/(ε₀A) = (1.0 × 10⁻⁸)/((8.85 × 10⁻¹²)(0.010)) = 1.1 × 10⁵ N/C.
4Near Earth's surface in fair weather, there is a downward electric field of about 100 V/m. What is the potential difference between your head (1.7 m) and your feet? Why don't you get a shock?
ΔV = Ed = (100)(1.7) = about 170 V. You don't get a shock because your body is a conductor: it becomes an equipotential and rearranges the field around you, and the atmosphere can supply only a tiny current.
5A small charged sphere (mass 2.0 × 10⁻⁴ kg) hangs between vertical plates 0.10 m apart with 2000 V across them. The thread makes an angle of 5.0° with the vertical. Find the sphere's charge.
E = 2000/0.10 = 2.0 × 10⁴ N/C. Balance: tan θ = qE/mg, so q = mg tan θ/E = (2.0 × 10⁻⁴)(9.8)(0.0875)/(2.0 × 10⁴) = 8.6 × 10⁻¹⁰ C.
Challenge
1Explain why the field outside a pair of oppositely charged parallel plates is nearly zero, using the field of a single large charged sheet.
A large flat sheet of charge makes a uniform field on both sides, pointing away from a + sheet and toward a − sheet. Between the plates, the + plate's field and the − plate's field point the same way (+ to −) and add. Outside, they point in opposite directions with equal size and cancel.
2A thundercloud's base is 1.0 km above flat ground, and the field between them is roughly uniform at 1.0 × 10⁵ V/m. Estimate the potential difference. How does this compare with the breakdown field of air (about 3 × 10⁶ V/m), and what does that suggest about how lightning starts?
ΔV = Ed = (1.0 × 10⁵)(1000) = 1 × 10⁸ V (100 million volts). The average field is about 30 times smaller than the breakdown value, so the air does not break down everywhere at once. Lightning starts where the field is locally much stronger (around charged regions, ice particles or sharp objects) and then grows step by step.
3Two parallel plates, 3.0 cm apart, are at 0 V and 150 V. A third, thin metal plate connected to nothing is inserted halfway between them, parallel to the others. What is its potential, and does it change the field in the gaps?
The thin conductor becomes an equipotential at the potential of the place it sits: 75 V. Charges on its two faces rearrange (−on the side facing the + plate, + on the other) but the field in each gap stays 5.0 × 10³ V/m (75 V over 1.5 cm), the same as before.
Quiz
1. Between two large, oppositely charged parallel plates, the electric field is
2. Plates 2.0 mm apart have 10 V across them. The field is
3. A unit equivalent to N/C is
4. A charge moves along a line parallel to the plates, between them. The work done by the electric field is
5. Plates are connected to a battery, and their separation is tripled. The field becomes
6. Which is the best gravitational comparison for the field between parallel plates?
Big idea (D2.3, D3.3): between parallel plates the field is uniform, E = ΔV/d, so a charge feels the same force everywhere in the gap, and the potential changes steadily across it. It is the electric version of the near-Earth gravitational field.
How to help: Ask your teen to draw the field and equipotential lines between plates, then to explain how it differs from a point charge's field. A good check: "If you pull the plates apart while the battery stays connected, what happens to E?" (It drops.)
Common difficulties:
- Forgetting to convert mm or cm to m in E = ΔV/d.
- Thinking the force is stronger near one plate.
- Mixing up the cases "battery connected" (ΔV fixed) and "battery disconnected" (Q fixed).
- Getting the force direction wrong for electrons (toward the + plate).
Safety: the foil capacitor activity uses only static charge from a balloon. Never connect home-made plates to a wall outlet.