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(b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor.

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Consider capacitor of capacitance C being charged from a battery of emf V. Initially, its two plates are uncharged. The positive charge is transferred from the second plate to the first plate in a very small amount of $dq$ till the first plate acquires the final charge +Q and the second plate acquires the charge â€“Q.

At a certain instant t, during the process of charging, let q be the charge on the plate and V is the potential difference between the plates.

Then $q = CV$

$V = \dfrac{q}{C}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

Where $C$ is the capacitance of the capacitor.

The amount of work done in transferring a small amount of charge $dq$ against the potential difference is,

$dw = Vdq$

Substitute equation (1) in above equation,

$dw = \dfrac{q}{C}dq$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (2)

The total work done in transferring charge Q is given by,

$W = \int\limits_0^W {dw} $

$W = \int\limits_0^W {\dfrac{q}{C}dq} $

$W = \dfrac{1}{C}\int\limits_0^W {qdq} $

Use this formula for integration,$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $

After integration we get,

$W = \dfrac{1}{C}\left[ {\dfrac{{{q^2}}}{2}} \right]_0^Q$

This is the amount of work done is stored as electrostatic potential energy U in the capacitor.

Thus energy stored in the capacitor is $U = \dfrac{1}{C}\left[ {\dfrac{{{Q^2}}}{2}} \right]$

$ = \dfrac{{{Q^2}}}{{2C}}$

We have $Q = CV$ substitute in above equation we get

$U = \dfrac{1}{2}C{V^2}$ These are the different kinds of equations of energy stored in a capacitor.

Energy density: energy stored per unit volume of the space between the plates is called energy density.

consider the separation $d$ of a parallel plate capacitor. Let us have $U = \dfrac{1}{2}C{V^2}$be the energy stored in the space of volume V=$A \times d$ between the plates.

Hence energy density, $u = \dfrac{{energy}}{{volume}}$

$ = \dfrac{{\dfrac{1}{2}C{V^2}}}{{Ad}}$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦( 3)

But the capacitance $C = \dfrac{{{\varepsilon _0}A}}{d}$ and potential difference $V = Ed$

Substituting in equation (3)

$u = \dfrac{1}{2}\dfrac{{{\varepsilon _0}A}}{d} \times \dfrac{{{E^2}{d^2}}}{{Ad}}$

$u = \dfrac{1}{2}{\varepsilon _0}{E^2}$

This is a general expression which is applicable for electric fields due to any configuration of charges.

Energy stored before connection,

${U_1} = \dfrac{{{Q^2}}}{{2C}}$

${U_2} = 2\dfrac{{{{\left( {\dfrac{Q}{2}} \right)}^2}}}{{2C}}$

Energy stored after connection,

$ = \dfrac{{{Q^2}}}{{4C}}$

${U_2} < {U_1}$

From this we can say, storage of energy in the combination is less than that of the single capacitor.

Charging a capacitor means transferring electrons from one plate to another.

The capacitance of a capacitor depends on (i) the shape of its plates, (ii) separation between plates, and (iii) the nature of dielectric material between the plates.

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