Parallel Plate Capacitor

1. Concept Overview

A parallel plate capacitor is the simplest and most widely studied form of a capacitor.
It consists of two large, flat, parallel conducting plates separated by a small distance d, with a dielectric medium (air/vacuum or another insulating material) between them.

When a potential difference is applied across the plates:

• One plate accumulates +Q charge,
• The other accumulates –Q,
• An electric field E is established between them.

A parallel plate capacitor is used to understand the basic idea of capacitance, because its geometry is simple and allows the electric field between plates to be uniform (except at edges).

The capacitance of a parallel plate capacitor is:

[C = \dfrac{\varepsilon A}{d}]

where

• [\varepsilon] [= \varepsilon_{0} \varepsilon_{r}] is the permittivity of medium,
• [A] is plate area,
• [d] is separation.

 

This model forms the foundation for understanding all real capacitors.

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2. Clear Explanation and Derivation

Electric Field Between Plates

For two large plates with surface charge densities [+\sigma] and [-\sigma],
the electric field due to each plate is:

[E = \dfrac{\sigma}{2\varepsilon_{0}}]

Between the plates, fields add:

[E = \dfrac{\sigma}{\varepsilon_{0}}]

For vacuum or air, [\varepsilon = \varepsilon_{0}].
For dielectric medium, [\varepsilon = \varepsilon_{0}\varepsilon_{r}].

Potential Difference

[V = Ed = \dfrac{\sigma d}{\varepsilon}]

Capacitance Derivation

Charge on each plate:

[Q = \sigma A]

Capacitance:

[C = \dfrac{Q}{V} = \dfrac{\sigma A}{\dfrac{\sigma d}{\varepsilon}}]

[C = \dfrac{\varepsilon A}{d}]

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3. Dimensions and Units

Dimensions of Capacitance

[C] = M^{-1}L^{-2}T^{4}A^{2}

SI Unit

• Farad (F)
• Practical units: μF, nF, pF

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4. Key Features

• Electric field between plates is uniform (ignoring edge effects).
• Capacitance increases with:
  • Larger plate area
  • Smaller separation
  • Higher dielectric constant
• Stores potential energy in the electric field:
  [U = \dfrac{1}{2}CV^{2}]
• Used in filters, tuning circuits, oscillators, flash circuits, DC blocking, etc.

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5. Important Formulas to Remember

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6. Conceptual Questions With Solutions

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7. FAQ / Common Misconceptions

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8. Practice Questions (With Step-by-Step Solutions)

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1. A parallel plate capacitor has area [0.5\ \text{m}^2] and separation [2 \times 10^{-3}\ \text{m}]. Find capacitance in air.

[C = \dfrac{\varepsilon_{0}A}{d}]

[C] [= \dfrac{8.85 \times 10^{-12} \times 0.5}{2 \times 10^{-3}}]

[C] [= 2.21 \times 10^{-9}\ \text{F}]

[\boxed{C = 2.21\ \text{nF}}]

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2. A capacitor has capacitance 10 pF in air. A dielectric of [\varepsilon_{r} = 5] is inserted. Find new capacitance.

[C’ = \varepsilon_{r}C]

[C’ = 5 \times 10\ \text{pF}]

[\boxed{50\ \text{pF}}]

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3. A capacitor with plate area [0.2\ \text{m}^2] and separation 5 mm stores charge [Q = 10\ \mu C]. Find voltage across it (air).

[C = \dfrac{\varepsilon_{0}A}{d}]

[C] [= \dfrac{8.85 \times 10^{-12} \times 0.2}{5 \times 10^{-3}}]

[C = 3.54 \times 10^{-10}\ \text{F}]

[V = \dfrac{Q}{C}]

[V] [= \dfrac{10 \times 10^{-6}}{3.54 \times 10^{-10}}]

[V] [\approx 2.82 \times 10^{4}\ \text{V}]

[\boxed{28.2\ \text{kV}}]

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4. A parallel plate capacitor is charged to 100 V. If [C = 2\ \text{nF}], find energy stored.

[U = \dfrac{1}{2}CV^{2}]

[U] [= \dfrac{1}{2} \times 2 \times 10^{-9} \times (100)^{2}]

[U = 1 \times 10^{-5}\ \text{J}]

[\boxed{10\ \mu J}]

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5. Doubling plate area and halving separation increases capacitance by what factor?

Initial: [C] [= \dfrac{\varepsilon A}{d}]
Final: [C’] [= \dfrac{\varepsilon (2A)}{\dfrac{d}{2}}] [= \dfrac{4\varepsilon A}{d}]

[\dfrac{C’}{C} = 4]

Capacitance increases by a factor of 4