Parallel Plate Capacitor

Two parallel conducting plates with equal and opposite charge densities create a uniform electric field between them and zero field outside. This configuration is fundamental to capacitors and many electrostatic devices.

Setup

  • Two large parallel conducting plates
  • Plate separation $d$ (small compared to plate dimensions)
  • Surface charge density $+\sigma$ on one plate, $-\sigma$ on the other

Electric Field Analysis

Each plate produces a field $E_{single} = \frac{\sigma}{2\varepsilon_0}$ (from Electric Field of Charged Plane).

Regions

Region A (outside, near positive plate):

  • Field from $+\sigma$ plate: points right (away from positive)
  • Field from $-\sigma$ plate: points left (toward negative)
  • Result: $E_A = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$

Region B (between plates):

  • Field from $+\sigma$ plate: points right
  • Field from $-\sigma$ plate: points right (toward negative plate)
  • Result: $E_B = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$

Region C (outside, near negative plate):

  • Field from $+\sigma$ plate: points right
  • Field from $-\sigma$ plate: points left
  • Result: $E_C = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$

Final Result

Region Electric Field
Between plates $E = \frac{\sigma}{\varepsilon_0}$
Outside plates $E = 0$

Key Properties

  • Uniform field between plates (constant magnitude and direction)
  • Zero field outside (fields cancel)
  • Field points from positive plate to negative plate
  • Field strength proportional to charge density

Alternative Forms

Using $Q = \sigma A$ (total charge on plate of area $A$):

$$E = \frac{Q}{\varepsilon_0 A}$$

Or in terms of voltage $V$ and plate separation $d$:

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

Example Calculation

Problem: Two parallel plates with area $A = 100$ cm² = $10^{-2}$ m² produce $E = 7.20 \times 10^3$ N/C between them. Find the charge on each plate.

Solution: $$E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$$

$$Q = \varepsilon_0 A E = (8.85 \times 10^{-12})(10^{-2})(7.20 \times 10^3)$$

$$Q = 6.37 \times 10^{-10} \text{ C} = 0.637 \text{ nC}$$

Applications

  1. Capacitors — Store electrical energy
  2. Deflection plates — In oscilloscopes, particle accelerators
  3. Uniform field region — For experiments requiring constant $E$
  4. Electrostatic precipitators — Remove particles from air

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