19.3 Capacitors | Electric Potential And Capacitor | Physics 12

What is a Capacitor?

A capacitor is a device that stores electric charge and energy. It consists of two conducting plates (or surfaces) separated by an insulating material called a dielectric (or vacuum/air).

Capacitance ( $C$ ) is the ability of a capacitor to store charge. It is defined as the ratio of the magnitude of charge $Q$ stored on either plate to the potential difference $V$ between the plates:

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

SI Unit: Farad (F)
1 Farad = 1 Coulomb per Volt (C/V)
In practice, capacitors are rated in microfarads ( $μ$ F) or picofarads (pF) since 1 F is very large.

Parallel Plate Capacitor

The simplest capacitor consists of two parallel conducting plates, each of area $A$ , separated by a distance $d$ .

Capacitance with Vacuum Between Plates

For a parallel plate capacitor with vacuum (or air) between the plates:

$C_{vac} = \frac{ϵ _{0} A}{d}$

where:

$ϵ_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}$ is the permittivity of free space
$A$ = area of each plate (m²)
$d$ = separation between plates (m)

Key relationships:

$C \propto A$ — larger plates store more charge
$C \propto \frac{1}{d}$ — closer plates increase capacitance

Worked Example

A parallel plate capacitor has plates of area $0.02 m^{2}$ separated by $1 mm = 1 \times 1 0^{- 3} m$ . Find its capacitance.

$C = \frac{ϵ _{0} A}{d} = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 0.02 )}{1 \times 1 0 ^{- 3}} = 1.77 \times 1 0^{- 10} F = 177 pF$

Effect of a Dielectric

A dielectric is an insulating material (e.g., glass, mica, paper) placed between the plates of a capacitor.

Electric Polarization

When a dielectric is placed in an electric field, its molecules align or distort — positive charges shift slightly in the direction of the field and negative charges shift opposite. This is called electric polarization. It creates induced surface charges on the dielectric that produce an internal electric field opposing the external field.

How Dielectric Increases Capacitance

The induced internal field reduces the net electric field between the plates, which lowers the potential difference $V$ for the same stored charge $Q$ . Since $C = Q / V$ , a lower $V$ means a higher capacitance.

With a dielectric of relative permittivity $ϵ_{r}$ (also called dielectric constant):

$C = ϵ_{r} C_{vac} = \frac{ϵ _{r} ϵ _{0} A}{d}$

Since $ϵ_{r} > 1$ for all dielectrics, inserting a dielectric always increases capacitance by a factor of $ϵ_{r}$ .

Effect of Changing Plate Separation (Connected to Battery)

If a capacitor remains connected to a battery (constant voltage $V$ ) and the plate separation $d$ is changed:

Since $C = \frac{ϵ _{0} A}{d}$ , doubling $d$ halves $C$ .
Energy stored: $U = \frac{1}{2} C V^{2}$ , so halving $C$ halves the energy stored.
Charge stored: $Q = C V$ , so halving $C$ also halves $Q$ .

Summary Table

Quantity	Formula	Effect of increasing $A$	Effect of increasing $d$	Effect of dielectric
Capacitance $C$	$\frac{ϵ _{r} ϵ _{0} A}{d}$	Increases	Decreases	Increases
Charge $Q$ (const. $V$ )	$C V$	Increases	Decreases	Increases
Voltage $V$ (const. $Q$ )	$Q / C$	Decreases	Increases	Decreases

What is a Capacitor?

A capacitor is a device that stores electric charge and energy. It consists of two conducting plates (or surfaces) separated by an insulating material called a dielectric (or vacuum/air).

Capacitance

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

SI Unit: Farad (F)
1 Farad = 1 Coulomb per Volt (C/V)
In practice, capacitors are rated in microfarads ( $μ$ F) or picofarads (pF) since 1 F is very large.

Parallel Plate Capacitor

The simplest capacitor consists of two parallel conducting plates, each of area $A$ , separated by a distance $d$ .

Capacitance with Vacuum Between Plates

For a parallel plate capacitor with vacuum (or air) between the plates:

$C_{vac} = \frac{ϵ _{0} A}{d}$

where:

$ϵ_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}$ is the permittivity of free space
$A$ = area of each plate (m²)
$d$ = separation between plates (m)

Key relationships:

$C \propto A$ — larger plates store more charge
$C \propto \frac{1}{d}$ — closer plates increase capacitance

Worked Example

A parallel plate capacitor has plates of area $0.02 m^{2}$ separated by $1 mm = 1 \times 1 0^{- 3} m$ . Find its capacitance.

$C = \frac{ϵ _{0} A}{d} = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 0.02 )}{1 \times 1 0 ^{- 3}} = 1.77 \times 1 0^{- 10} F = 177 pF$

Effect of a Dielectric

A dielectric is an insulating material (e.g., glass, mica, paper) placed between the plates of a capacitor.

Electric Polarization

How Dielectric Increases Capacitance

With a dielectric of relative permittivity $ϵ_{r}$ (also called dielectric constant):

$C = ϵ_{r} C_{vac} = \frac{ϵ _{r} ϵ _{0} A}{d}$

Since $ϵ_{r} > 1$ for all dielectrics, inserting a dielectric always increases capacitance by a factor of $ϵ_{r}$ .

Effect of Changing Plate Separation (Connected to Battery)

If a capacitor remains connected to a battery (constant voltage $V$ ) and the plate separation $d$ is changed:

Since $C = \frac{ϵ _{0} A}{d}$ , doubling $d$ halves $C$ .
Energy stored: $U = \frac{1}{2} C V^{2}$ , so halving $C$ halves the energy stored.
Charge stored: $Q = C V$ , so halving $C$ also halves $Q$ .

Summary Table

Quantity	Formula	Effect of increasing $A$	Effect of increasing $d$	Effect of dielectric
Capacitance $C$	$\frac{ϵ _{r} ϵ _{0} A}{d}$	Increases	Decreases	Increases
Charge $Q$ (const. $V$ )	$C V$	Increases	Decreases	Increases
Voltage $V$ (const. $Q$ )	$Q / C$	Decreases	Increases	Decreases