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Electrostatics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectrostatics5 Marks
Question
Define capacitance of a capacitor. Derive expression for stored energy between plates of parallel plate capacitor. Show that energy-density between plates of parallel capacitor can be expressed as $\frac{1}{2}\in_{0}E^{2}$ , when E = Electric Field between plates.
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Answer

Capacitance: It is the capacity of a conductor for acquiring electric charge. It is defined as follows :
"The electrical capacitance of a conductor is defined as the ratio of the charge given to the conductor and potential raised due to it." It is denoted by C.
If $q$ be the quant ity of charge given to any conductor and V be the potential raised by the charge, then :
$\text { Capacitance } =\frac{\text { ElectricCharge }}{\text { RiseinPotential }} $
$\Rightarrow \quad C =\frac{q}{V}$ ...(1)
It depends upon the size and shape of the conductor and upon the nature of surrounding medium and presence of other conductors.
If $V =1$, then from equation (1), we get
$C=q .$
Thus, the capacitance of a conductor is numerically equal to the charge to be given to a conductor to raise its potential by unity.
S.I. Unit of capacitance. The SI unit of capacitance is farad. It is so named in the honour of Michael Faraday and is depicted by F.
Energy of a charged capacitor : The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential.Let $q$ be the charge given to the capacitor so that its potential becomes V , then :
$V=\frac{q}{C}$
where C is the capacitance of the capacitor, $1 / C$ is a constant here.
$\therefore \quad V \propto q .$
Graph between potential and charge is shown in the following figure. It is a straight line OA. Area OAB represents the energy stored in the capacitor in charging it. Thus stored potential energy \
$U =$ Area of $\Delta OAB$
$=\frac{1}{2} OB \times BA $
$=\frac{1}{2} \times q \times \frac{q}{C} $
$\Rightarrow U =\frac{ 1 }{2}\left(\frac{ q ^2}{ C }\right) $ ...(1)
$\text { But } q =CV $
$\therefore U =\frac{1}{2} \frac{(CV)^2}{C} \Rightarrow U =\frac{ 1 }{2} C V ^2 $ ...(2)
$\Rightarrow U =\frac{1}{2} CV(V) ; \text { but } CV=q $
$\therefore U =\frac{ 1 }{2} q V$ ...(3)
This energy of the charged capacitor remains in the medium between plates.
Energy density : The energy of a charged parallel plate capacitor is stored in the medium existing between its plates. "Energy density of a parallel plate capacitor is defined as the amount of energy stored per unit volume of the space (i.e. medium) between the plates of the capacitor." It is denoted by $u$.
Image
i.e. $\quad u=\frac{\text { Total Energy Stored ' } U}{\text { Volume of space between the plates }}$
But $\quad U =\frac{1}{2} CV ^2$ and Volume $= A \times d$
where
$C =$ capacitance of the capacitor,
$V =$ potential difference between the plates of the capacitor,
$A=$ area of each of the plates of the capacitor and
$d=$ distance between the plates of the capacitor.
$\therefore \quad u=\frac{\frac{1}{2} CV^2}{A \times d} \Rightarrow u=\frac{CV^2}{2 A d}$
But capacitance of the air capacitor
$C =\frac{\varepsilon_0 A}{d} $
$\Rightarrow \quad u =\frac{\left(\varepsilon_0 A / d\right) \times V^2}{2 A d} \Rightarrow u =\frac{1}{2} \varepsilon_0\left(\frac{ V }{ d }\right)^2$ ... (1)
But $\frac{V}{d}=$ Electric field intensity between the plates of the capacitor = E
$\therefore \quad u =\frac{1}{2} \varepsilon_0 E ^2$ ... (2)
If the space between the plates be filled with some medium of dielectric constant $K$, then the energy density would be
$u=\frac{1}{2} K \varepsilon_0 E^2$ ...(3)
The S. I. unit of $u$ is $\text { joule/metre}^3$.

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