Parallel Plate Capacitor: Uniform Field

Two parallel plates with equal and opposite charges create a remarkably simple electric field: uniform (constant magnitude and direction) in the region between the plates, and (ideally) zero outside. This is the geometry inside every capacitor and the principle behind cathode-ray tubes, particle accelerators, and electrostatic precipitators.

The Field

For plates with surface charge density $\sigma$ (charge per unit area) on each, the field between them is:

$E = \dfrac{\sigma}{\varepsilon_0}$

where $\varepsilon_0 = 8.854 \times 10^{-12}\;\text{C}^{2}/(\text{N}\cdot\text{m}^{2})$ is the permittivity of free space.

If you know the voltage $V$ across a plate gap of width $d$, the field is also:

$E = \dfrac{V}{d}$

This is a clean linear relationship — doubling the voltage (for fixed $d$) doubles the field; doubling the gap (for fixed $V$) halves it.

Step-by-Step Solution

Given: Two parallel plates with $V = 100\;\text{V}$ across a gap of $d = 0.01\;\text{m}$ (1 cm).

Find: The electric field, the force on a $+1\;\text{nC}$ test charge, and its acceleration if it has mass $1\;\text{μg} = 10^{-9}\;\text{kg}$.

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Step 1 — Compute the field.

$E = \dfrac{V}{d} = \dfrac{100}{0.01} = 10{,}000\;\text{V/m}$

The field points from the positive plate to the negative plate, with magnitude 10 kV/m everywhere in the gap.

Step 2 — Force on the test charge.

$F = qE = (10^{-9})(10{,}000) = 10^{-5}\;\text{N}$

A small force, but on a small charge.

Step 3 — Acceleration of the test charge.

$a = \dfrac{F}{m} = \dfrac{10^{-5}}{10^{-9}} = 10^{4}\;\text{m/s}^{2}$

That's about 1000 g of acceleration. Even tiny electrostatic forces can produce huge accelerations on light particles — that's the basis of electron-beam steering in old CRT TVs.

Step 4 — Capacitance and stored charge.

The relationship between charge, capacitance, and voltage is:

$Q = CV$

where the capacitance depends on the plate area $A$ and the gap:

$C = \dfrac{\varepsilon_0 A}{d}$

For $A = 1\;\text{cm}^{2} = 10^{-4}\;\text{m}^{2}$ and $d = 0.01\;\text{m}$:

$C = \dfrac{(8.854 \times 10^{-12})(10^{-4})}{0.01} \approx 8.85 \times 10^{-14}\;\text{F} = 0.0885\;\text{pF}$

A tiny capacitance. At 100 V:

$Q = CV \approx (8.85 \times 10^{-14})(100) \approx 8.85 \times 10^{-12}\;\text{C} \approx 8.85\;\text{pC}$

Step 5 — Energy stored in the capacitor.

$U = \tfrac{1}{2}CV^{2} = \tfrac{1}{2}(8.85 \times 10^{-14})(10^{4}) \approx 4.4 \times 10^{-10}\;\text{J}$

About 0.44 nanojoules — modest, because the capacitance is small.

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Answer: A 100 V battery across a 1 cm plate gap creates a uniform field of 10{,}000 V/m between the plates. A 1 nC test charge feels a 10 μN force; if it weighs 1 microgram, it accelerates at about $10^{4}\;\text{m/s}^{2}$ — over 1000 g.

The field outside the plates is essentially zero (apart from "fringe field" near the edges, which we ignore in the ideal case).

Try It

• Adjust the voltage $V$ — the field scales linearly.
• Adjust the gap $d$ — the field scales inversely.
• The field lines between the plates are evenly spaced and parallel, indicating the uniform field. Outside the plates, no lines.