# Energy Stored in a Charged Capacitor, Energy Density and Loss of Energy due to Joining of Capacitor

#### Energy Stored in a Charged Capacitor

In order to charge a capacitor certain work is done against the electrostatic force of repulsion, This work done is stored in the form of electric energy of the capacitor.

Suppose a capacitor having capacitance 'C' is being charged with the help of the cell. If 'V' represents the potential difference the small work done 'dW' on storing the charge 'dQ' is given by

$$dW = dQ. V$$

But Q= CV or v = Q/C

Using this in equation (i), we get

$$\text {or,} dW = \frac {Q}{C}.dQ $$

Hence, the total work done'W' on storing charge 'Q' on such capacitor is obtained by integrating 'dW' between the limits zero and 'Q'.

$$\text {or,} W = \int dW $$

$$\text {or,} W = \int _0^Q\frac QC dQ $$

$$\text {or,} W = \frac {1}{C} \int _0^Q Q dQ $$

$$\text {or,} W = \frac {1}{C} \left [\frac {Q^2}{2} \right ]_0^Q $$

$$\text {or,} W = \frac {1}{C} \left [\frac {Q^2 - 0^2}{2} \right ]$$

$$\text {or,} W = \frac {1}{1} \frac {Q^2}{C} $$

The work done is stored in the form of electrical energy of the capacitor.

$$\therefore \text {Energy stored} (U) = \text {Work done} (W)$$

$$ U = \frac 12\frac {Q^2}{C} $$

Also \(Q = CV\)

$$ \text {or,} U = \frac 12\frac {(CV)^2}{C} $$

$$ \text {or,} U = \frac 12 CV^2$$

$$Q = C.V$$

$$ \text {or,}U = \frac 12 (CV).V$$

$$\therefore U = \frac {1}{2} QV $$

$$\text {or,} U = \frac {1}{2} \frac {Q^2}{C}=\frac 12 C.V^2 =\frac {1}{2} QV $$

This is the expression for energy stored (U) in different form on a charged capacitor.

#### Energy Density (u)

The energy density of a capacitor is the energy stored per unit volume. It is denoted by u. Suppose, A is the plate area and 'd' is the distance between the plates. Then,

$$\text Volume of capacitor= A.d $$

$$Energy Density (u) = \frac {\text {energy}}{\text {volume}} $$

$$= \frac{\frac {1}{2} CV^2}{A.d} $$

$$\frac 12 \frac {CV^2}{Ad} $$

But for a parallel plate capacitor capacitance \( (C) = \frac {\epsilon _o. A}{d}\)

using this in above equation, we get

$$\text {or,} u = \frac 12 \frac {\epsilon _o. A}{d}\frac {V^2}{A.d}$$

$$= \frac 12 \epsilon_o \left (\frac Vd \right )^2$$

But, we know

$$\frac Vd = E $$

$$ \text {or,} u = \frac 12 \epsilon _o E^2 $$

This is the expression for energy density of a charged capacitor having no dielectric medium between two poles.

If a dielectric medium is present between the plates. Then,

$$ u = \frac 12 \epsilon E^2 $$

**Loss of Energy due to joining of Capacitors**

Suppose two capacitors 'C_{1}' and'C_{2}' are charged up to 'Q_{1}' and'Q_{2}'. So that, the potential difference becomes'V_{1}' and'V_{2}' respectively.

$$\therefore \text {Total energy stored} (u) =\frac 12 C_1V_1^2 +\frac 12 C_2V_2^2\dots (i)$$

Now, similar plates of the charged capacitor are joined and redistribution of charge takes place until a common potential difference (V) is maintained across the combination.

Here, the charge present on capacitor 'C_{1}' and'C_{2}' becomes 'Q_{1}' and'Q_{2}' respectively.

Then, from conservation of charge

$$ Q_1 + Q_2 = Q_1' + Q_2' $$

$$C_1V_1 + C_2V_2 = C_1V + C_2V$$

$$V = \frac {C_1V_1 + C_2V_2}{C_1 + C_2}$$

This gives the common potential difference (P.d) across the combination after joining the similar plates of charged capacitors.

Then, total energy stored after joining similar plate\( (u') = \frac 12 C_1V^2 + \frac 12 C_2V^2 \)

$$=\frac 12 (C_1 + C_2)V^2$$

$$=\frac 12 (C_1 + C_2) \left (\frac {C_1V_1 + C_2V_2}{C_1 + C_2} \right )^2 $$

$$=\frac 12 \frac {(C_1V_1 + C_2V_2)^2}{(C_1 + C_2)}$$

$$\therefore \text {loss of energy} (\Delta U) = U - U' $$

$$=\frac 12 C_1V_1^2 +\frac 12 C_2V_2^2 -\frac 12 \frac {(C_1V_1 + C_2V_2)^2}{(C_1 + C_2)}$$

$$= \frac {1}{2(C_1 + C_2)} (C_1 + C_2)C_1 V_1^2 + (C_1 + C_2) C_2V_2^2 - (C_1 V_1^2 +2C_1 C_2V_1V_1- C_2V_2^2)^2$$

$$= \frac {C_1^2 V_1^2 + C_1C_2V_1^2+ C_1C_2V_2^2 + C_2^2V_1^2 - C_1^2V_1^2 -2C_1C_2V_1V_2 - C_2^2V_2^2} {2(C_1 + C_2)} $$

$$= \frac {C_1C_2 (V_1^2 + V_2^2 -2V_1V_2)}{2(C_1 + C_2)}$$

$$\Delta U = \frac {C_1C_2}{2(C_1 + C_2)} (V_1 -V_2)^2$$

Whatever be the value of'V_{1}' and'V_{2}'. (V_{1}'- V_{2}') is always positive. Hence, joining of similar plates of the charged capacitors always accompanies with loss of energy. This energy loss appears in the front of light, heat and sound energy etc.

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