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Maharashtra State BoardHSC Science (Computer Science) 12th Standard Board Exam
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पाठ्यक्रम

The Parallel Plate Capacitor

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Estimated time: 5 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

Related QuestionsVIEW ALL [29]

A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.

A ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A if the refractive index of the material of the sphere is \[\sqrt{3}\].

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

The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates.

Solve the following question.
A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single 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.]

A parallel-plate capacitor with plate area 20 cm2 and plate separation 1.0 mm is connected to a battery. The resistance of the circuit is 10 kΩ. Find the time constant of the circuit.

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