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The Parallel Plate Capacitor

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Estimated time: 8 minutes
CBSE: Class 12

Definition: The Parallel Plate Capacitor

A capacitor that consists of two large, parallel, conducting plates separated by a small distance is called a parallel plate capacitor.

CBSE: Class 12

Formula: Capacitance of a Parallel Plate Capacitor

For two plates separated by distance d:

\[C=\frac{\varepsilon_0A}{d}\]

With a dielectric medium:

\[C=\frac{K\varepsilon_0A}{d}\]

CBSE: Class 12

Step-by-Step Derivation of Capacitance

Step 1: Surface Charge Density

σ = \[\frac {Q}{A}\]

Step 2: Electric Field Between the Plates

Using the result from Gauss's Law for a large sheet of charge, the electric field due to each plate is \[\frac {σ}{2ε_0}\]. For two plates with opposite charges, both contributions add between the plates:

E = \[\frac {σ}{ε_0}\] = \[\frac {Q}{ε_0A}\]

Outside the plates, the fields are equal and opposite → they cancel → E = 0 outside.

Step 3: Potential Difference Between the Plates

V = E ⋅ d = \[\frac {Q⋅d}{ε_0A}\]

Step 4: Capacitance

C = \[\frac {Q}{V}\] = \[\frac{Q}{\frac{Qd}{\varepsilon_0A}}\]
 
Result:
C = \[\frac {ε_0A}{d}\]

Capacitance depends only on geometry (A and d) and the medium — NOT on Q or V.

CBSE: Class 12

Factors Affecting Capacitance

Factor Effect on Capacitance (C) Reason
Increase plate area (A) Capacitance increases (C ∝ A) A larger plate area allows more charge to be stored for the same potential difference.
Increase plate separation (d) Capacitance decreases ((C ∝ 1/d)) Greater separation reduces the electric field interaction between the plates.
Insert a dielectric (K > 1) Capacitance increases (C = Kε0A/d) The dielectric reduces the effective electric field, enabling more charge storage.
Change charge (Q) or potential difference (V) No effect on capacitance Capacitance depends only on the geometry of the conductors and the dielectric medium.
CBSE: Class 12

Effect of Dielectric

When a dielectric slab of dielectric constant K is inserted between the plates:

Formula with Dielectric:

C = \[\frac {Kε_0A}{d}\]
  • The electric field is reduced to E′ = \[\frac {E}{K}\] due to polarisation of the dielectric
  • Since V′ = E′ ⋅ d decreases, and C = Q/V′, the capacitance increases by factor K
CISCE: Class 12

Working Principle / Energy Storage

  • A parallel plate capacitor consists of two conducting plates separated by a dielectric (or vacuum), and stores electric charge when a potential difference is applied across them.
  • Connecting the plates to a voltage source causes one plate to accumulate positive charge (+Q) and the other negative charge (−Q), creating a uniform electric field between the plates.
  • This charge separation stores electrical energy in the field between the plates, which can later be released back into a circuit when needed.
  • The amount of charge stored depends on the capacitance (C = ε₀A/d), meaning larger plate area or smaller plate separation allows more energy to be stored for the same voltage.
CISCE: Class 12

Applications

  • Filters and oscillators: used in electronic circuits to control signal frequency and smooth out voltage fluctuations.
  • Energy storage in power supplies: store and release charge to stabilise output voltage.
  • Touchscreen and pressure sensors: detect changes in capacitance caused by touch or pressure to register input.
  • Radio and TV tuning circuits: used to select specific frequencies by adjusting capacitance in tuning circuits.

Related QuestionsVIEW ALL [29]

Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery

A parallel plate air condenser has a capacity of 20µF. What will be a new capacity if:

1) the distance between the two plates is doubled?

2) a marble slab of dielectric constant 8 is introduced between the two plates?

Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.

Define the capacitance of a capacitor and its SI unit.

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Draw a neat labelled diagram of a parallel plate capacitor completely filled with dielectric.

Answer the following question.
Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.

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