Sharing of Charges between two Capacitors and Dielectric

Sharing of Charges between two Capacitors

Consider two charged capacitors of capacitance C₁ and C₂ having potentials V₁ and V₂ respectively as shown in the figure. Then, initially the charge on capacitor C₁ is Q₁ = C₁V₁ and on the charge onC₂is Q₂ = C₂V₂.

Suppose two capacitors are connected to each other with a wire with like charge at the same point. The capacitors will share with each other till they acquire a common potential V. Let Q₁' and Q₂' be the final charges on the respective capacitors. Then the final charge on C₁,

$$Q_1 = C_1V \dots (i)$$

and the final charge onC₂,

$$Q_2 = C_2V \dots (ii)$$

Since the charged is conserved, the initial total charge of the two capacitors is equal to their final charge. So,

$$\text {or,} Q_1 + Q_2 = Q_1' + Q_2'$$

$$\text {or,} C_1V_1 + C_2V_2 = C_1V + C_2V$$

$$ \text {or,} V = \frac {C_1V_1 + C_2V_2 }{C_1 + C_2} \dots (iii)$$

This is the common potential obtained on combining the capacitors. The final charge stored in C₁ and C₂are

$$\text {or,} Q_1' = C_1\left (\frac {C_1V_1 + C_2V_2 }{C_1 + C_2} \right ) \dots (iv)$$

$$\text {or,} Q_2' = C_2\left (\frac {C_1V_1 + C_2V_2 }{C_1 + C_2} \right ) \dots (v)$$

Before connection, if Q be the total charge, then

$$\text {or,} Q = C_1V_1 + C_2V_2$$

Therefore, equation (iv) and (v) becomes

$$Q_1' = \frac {C_1}{C_1 +C_2}Q$$

$$Q_2' = \frac {C_2}{C_1 +C_2}Q$$

Uses of Capacitors

Capacitors can be used as device for storing charge.
They are used in scientific investigation.
They are used in increasing the efficiency of alternating current power transmission.
They give an electric field of desired configuration.
The ignition system of any automobile engine contains a capacitor to eliminate the sparking of the points when they open or close.
They are used in oscillators, in filter circuits, time delay device and many other electronic devices.

Action of dielectric on a capacitor

When a capacitor is fully charged it cannot accommodate furthermore charge. But when a dielectric medium is introduced between its plates, its molecules get polarized and hence enable the capacitor to accommodate more charge. This causes the capacitance of the capacitor to be increased. If ‘C

Consider a parallel plate capacitor each plate of area A and separated by a distance d apart. If ‘C_o’ represents the capacitance of a capacitor without any medium present between its plates and ‘C_m’ represents the capacitance when a dielectric material having dielectric constant ‘K’ is introduced then,

$$C_o = \frac {\epsilon_o A}{d}\dots (i)$$

$$C_m = \frac {\epsilon A}{d} =\frac {K \epsilon _o A}{d} \dots (ii)$$

Dividing equation (ii) by (i)

$$\frac {C_m}{C_o} = \frac {K \epsilon _o A}{d}\times\left (\frac {d} {\epsilon_o A} \right )$$

$$C_m = KC_o$$

As the dielectric constant K for any medium is greater than that of Vacuum or air, so

$$\boxed {C_m >C_o}$$

Hence, the action of dielectric on a capacitor is to increase in capacitance.

Relative Permittivity (Dielectric constant)

The ratio of the capacitance of a capacitor having a dielectric between the plates to the capacitance of the same capacitor without any dielectric is called the relative permittivity or dielectric constant of the material used. The relative permittivity of the substance is denoted by ε_r.

$$\epsilon_r = \frac {\text {capacitance of given capacitor with space between the plates fillled with dielectric}}{\text{capacitance of same capacitor with plates in vacuum}}$$

$$\text {or,} \frac {C_m}{C_o} $$

The capacitance of a parallel plate capacitor is $$$$

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

If the medium between the plates in vacuum or air, then

$$C_o = \frac {\epsilon_o A}{d}$$

Then, $\epsilon _r = \frac {C_m}{C_o} = \frac {\epsilon A/d}{\epsilon_o A/d} = \frac {\epsilon }{\epsilon _o}$

$$ \boxed {\therefore\epsilon= \frac {\epsilon }{\epsilon _o}} $$

The relative permittivity or dielectric constant is the ratio of the permittivity of the substance to that of free space. It has no dimension and unit.