Class 11 > Unit # 09: Capacitors > Capacitance & Parallel Plate Capacitor

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Capacitor and Capacitance

Video Lecture

Watch the complete video lecture below to understand capacitors, capacitance, and the step-by-step derivation for a parallel plate capacitor.

Capacitance and Capacitors

Capacitance

The potential $V$ of a conductor depends on its own charge $q$ as well as the charges on neighboring bodies. For an isolated conductor:

$$q \propto V$$

$$q = C V$$

$$C = \frac{q}{V}$$

"Where $C$ is the capacity of holding electric charge, or capacitance, which is defined as: The charge stored in a conductor per unit potential difference is known as the capacitance of that conductor."

The capacitance depends on the size and shape of the conductor.

Unit of Capacitance

Its SI unit is the Farad (F), which is defined as: "If by giving 1 coulomb charge to a conductor, the potential difference becomes 1 volt, then its capacitance will be 1 Farad."

Since the Farad is a very large unit, its sub-multiples are commonly used:

• $1\,\mu\text{F} = 10^{-6}\,\text{F}$
• $1\,\mu\mu\text{F} = 1\,\text{pF} = 10^{-12}\,\text{F}$

Capacitor

"A capacitor is a device used for storing electric charges. A capacitor consists of two conductors separated by air or any insulating material. The conductors carry equal and opposite charges. A capacitor is designed to have a large capacity for storing electric charge without having excessively large physical dimensions."

Capacitance of a Parallel Plate Capacitor

A parallel plate capacitor consists of two parallel metal plates separated by a small distance. The charges on each plate are uniformly distributed on the inner sides of the plates due to the attraction between opposite charges. The electric field $E$ between the plates is uniform except near its outer boundaries.

If $\sigma$ is the surface charge density, then:

$$\sigma = \frac{q}{A}$$

The electric field $E$ between the parallel plates in terms of surface charge density and permittivity ($\varepsilon_0$) is given by:

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

If $V$ be the potential difference between the plates separated by distance $d$, then the electric field can also be expressed as:

$$V = E \cdot d$$

Substituting $E = \frac{\sigma}{\varepsilon_0}$ into the potential equation:

$$V = \left(\frac{\sigma}{\varepsilon_0}\right) d$$

Since $\sigma = \frac{q}{A}$:

$$V = \frac{q \cdot d}{\varepsilon_0 A}$$

The capacitance $C$ of the capacitor is given by $C = \frac{q}{V}$. Substituting the expression for $V$:

$$C = \frac{q}{\left(\frac{q \cdot d}{\varepsilon_0 A}\right)}$$

$$C_{air} = \frac{\varepsilon_0 A}{d} \quad \text{--- (i)}$$

(When air is present between the plates)

Where:

• $A$ = Area of each plate
• $\varepsilon_0$ = Permittivity of free space
• $d$ = Distance between the plates of the capacitor

In the Presence of a Dielectric Medium

If the space between the plates is filled with an insulating material (dielectric medium) of relative permittivity $\varepsilon_r$, the absolute permittivity $\varepsilon$ becomes:

$$\varepsilon = \varepsilon_0 \varepsilon_r$$

Therefore, the capacitance in the presence of a dielectric medium ($C_m$) becomes:

$$C_m = \frac{\varepsilon A}{d}$$

$$C_m = \frac{\varepsilon_0 \varepsilon_r A}{d}$$

$$C_m = \varepsilon_r \left(\frac{\varepsilon_0 A}{d}\right)$$

$$C_m = \varepsilon_r C_{air}$$

The introduction of a dielectric medium increases the capacitance of a parallel plate capacitor by a factor equal to the relative permittivity ($\varepsilon_r$) of the medium.

Fig: Parallel plate capacitor carrying charges $+q$ and $-q$ separated by distance $d$.

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