# Parallel Plate Capacitor | Energy Stored in a Capacitor | Energy Density

### Capacitor:

Capacitor is a device for storing electric charge. It consists of a pair of conductors carrying equal and opposite charges (generally). Magnitude of this charge is known as the charge on the capacitor. Potential difference (V) between the two conductors (known as the potential across the capacitor) is proportional to the charge on the capacitor (Q)

Q ∝  V

Q = C V

Here the proportionality constant C is known as the capacitance of the capacitor. The value of capacitance depends on the geometry of the two conductors, their relative position and the medium between them.

## Parallel plate capacitor:

Electric field between the plates : Electric field due to the positive plate,

$\displaystyle E_1 = \frac{\sigma}{2 \epsilon_0} = \frac{Q}{2 \epsilon_0 A}$ ; from positive to negative plate.

Electric field due to the negative plate,

$\displaystyle E_2 = \frac{\sigma}{2 \epsilon_0} = \frac{Q}{2 \epsilon_0 A}$ ;  towards negative plate.

Net Field , $\displaystyle \vec{E} = \vec{E_1} + \vec{E_2}$

$\displaystyle E = \frac{Q}{2 \epsilon_0 A} + \frac{Q}{2 \epsilon_0 A}$

$\displaystyle E = \frac{Q}{ \epsilon_0 A} = \frac{\sigma}{\epsilon_0}$

Where σ = Q/A = Surface Charge density

P.d across the plates :

$\displaystyle V = E d = \frac{Q}{ \epsilon_0 A} d$

Capacitance $\displaystyle C = \frac{Q}{V} = \frac{Q}{ \frac{Q}{ \epsilon_0 A} d }$

Capacitance $\displaystyle C = \frac{ \epsilon_0 A}{ d }$

If Space between the plates is filled with dielectric medium of relative permittivity εr  or K then ,

Capacitance $\displaystyle C_m = \frac{ \epsilon_0 \epsilon_r A}{ d }$

Or , $\displaystyle C_m = \frac{ \epsilon_0 K A}{ d }$

### Capacitance of a parallel plate capacitor is :

(i) directly proportional to the area of the plates and

(ii) inversely proportional to the distance of separation between them

### Force acting between the plates:

Force acting between the plates = (Charge on one plate) x (Electric field due to the other plate)

$\displaystyle F = Q (\frac{Q}{ 2 \epsilon_0 A } )$

$\displaystyle F = \frac{Q^2}{ 2 \epsilon_0 A }$

### Energy stored in a capacitor :

Energy stored in a capacitor (U) = Amount of work done in charging the capacitor from initial uncharged state to the given charge state.
For a parallel plate capacitor it is the work done in increasing the separation of the charged plates from zero to d.

$\displaystyle W = \int_{0}^{d} F dx$

$\displaystyle = \frac{Q^2 d}{2 \epsilon_0 A}$

$\displaystyle U = \frac{Q^2 }{2 C}$

$\displaystyle U = \frac{1}{2 } C V^2$

$\displaystyle U= \frac{1}{2 } Q V$

The Energy stored per unit volume between the plates of capacitor is called Energy density .

$\displaystyle u = \frac{U}{volume}$

$\displaystyle U = \frac{1}{2 } C V^2$

Capacitance $\displaystyle C = \frac{ \epsilon_0 A}{ d }$ and V = E d

$\displaystyle U = \frac{1}{2 }(\frac{ \epsilon_0 A}{ d }) (E d)^2$

$\displaystyle u = \frac{U}{A d} = \frac{1}{2 }\epsilon_0 E^2$

### Effect of Dielectric :

Suppose the space between parallel plates capacitor is filled with dielectric having dielectric constant K , the electric field polarises the dielectric so that induced charges +Qin and -Qin appear in dielectric .

Due to dielectric , the electric field between the plates becomes

$\displaystyle E = \frac{E_0}{K}$ ; Where $\displaystyle E_0 = \frac{Q}{\epsilon_0 A}$

Capacitance of the Capacitor , $\displaystyle C = \frac{K \epsilon_0 A}{d}$

Energy Density , $\displaystyle u = \frac{1}{2 }K \epsilon_0 E^2$

Induced Charge , $\displaystyle Q_{in} = Q(1 – \frac{1}{K} )$

#### Capacitance of a Parallel Plate Capacitor filled partially with a dielectric :

$\displaystyle C = \frac{\epsilon_0 A}{d – t + \frac{t}{K}}$
Solved Example : A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?

Sol. Co =  8 pF = 8 × 10-12 F , K = 6

In first case , $\displaystyle C_0 = \frac{\epsilon_0 A}{d}$

By Formula , $\displaystyle C_K = \frac{K \epsilon_0 A}{d}$

In 2nd case , $\displaystyle C_K = \frac{K \epsilon_0 A}{d/2} = \frac{2 K \epsilon_0 A}{d}$

On dividing , $\displaystyle \frac{C_K}{C_0} = 2 K$

$\displaystyle C_K = 2K C_0$

CK = 2 × 6 × 8 × 10-12 = 96pF

Solved Example:  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 ?

Sol. A = 6 × 10–3 m2

d = 3 × 10–3 m , V =100 volt

$\displaystyle C_0 = \frac{\epsilon_0 A}{d}$

$\displaystyle C_0 = \frac{8.85\times 10^{-12}\times 6\times 10^{-3}}{3\times 10^{-3}}$

On Solving , Co = 17.7 × 10–12 F

Co = 17.7 pF

and q = CoV

q = 17.7 × 10–12 × 100

q = 1.77 × 10–9 C

Solved Example: Two identical sheets of a metallic foil are separated by d and capacitance of the system is C and charged to a potential difference E . Keeping the charge constant, the separation is increased by ‘ l ’. Then the new capacitance and potential difference will be :

(a) $\large \frac{\epsilon_0 A}{d}$ , E

(b) $\large \frac{\epsilon_0 A}{d + l}$ , E

(c) $\large \frac{\epsilon_0 A}{d+l}$ , $\large (1 + \frac{l}{d})E$

(d) $\large \frac{\epsilon_0 A}{d}$ , $\large (1 + \frac{l}{d})E$

Solution : q = CV = C1V1 where $\large C = \frac{\epsilon_0 A}{d}$

$\large q = \frac{\epsilon_0 A E}{d} = \frac{\epsilon_0 A E_1}{d+l}$

$\large E_1 = (\frac{d+l}{d})E$

$\large E_1 = (1+\frac{l}{d})E$

and , $\large C_1 = \frac{\epsilon_0 A}{d+l}$

Hence (C) is correct.

Solved Example : A capacitor of capacitance C is connected to a cell of emf V and when fully charged, it is disconnected. Now the separation between the plates is doubled. The change in flux of electric field through a closed surface enclosing the capacitor is

(a) Zero

(b) $\large \frac{C V}{\epsilon_0}$

(c) $\large \frac{C V}{2 \epsilon_0}$

(b) $\large \frac{ 2 C V}{\epsilon_0}$

Solution: $\large Flux = \frac{q_{in}}{\epsilon_0}$

The two plates of the capacitor have equal and opposite charges.

Hence, net charge enclosed by the given surface = 0

Flux is zero in both cases.

Hence change in flux = 0.

Hence (A) is correct.

Solved Example : A 12pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor ?

Sol. C = 12 × 10–12 F , V = 50 V

Electrostatic energy , $\displaystyle U = \frac{1}{2}CV^2$

$\displaystyle U = \frac{1}{2}(12\times 10^{-12})(50)^2$

U = 1.5 × 10–8 joule