Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • (a)    Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor. (b) A parallel plate ca

(a)    Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.

(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.

 

 

 

 
 
 
 
 

Answers (1)

(a)Let us consider, an uncharged capacitor of capacitance 'C' which is connected with a battery of e.m.f 'E' . Potential Difference across the capacitor will build up. Such that, the plate with positive terminal of the battery will lose electrons and the other plate will gain the electron. There is no transfer of charge in between the plates.

When the emf of the battery becomes equal to the potential difference across the capacitor, then the capacitor would be fully charged and will act as an open switch.

Derivation :

Let the charge on the capacitor is increased by dq amount then,

Then, the work done:-

dU=Vd{Q}'=\frac{{Q}'}{C}\; d{Q}'

On integrating it, we have

\int_{O}^{U}dU=\frac{1}{C}\int_{O}^{Q}{Q}'d\;{Q}'

U=\frac{1}{2}\left [ \frac{Q^{2}}{C} \right ]^{Q}_{O}

U=\frac{1}{2}\frac{Q^{2}}{C}

We know, C=\frac{Q}{V}

            U=\frac{1}{2}CV^{2}

            U=\frac{1}{2}QV

\therefore This is the work done is stored as energy on the capacitor 

(b)     We know, the energy stored by the capacitor when it is charged is given as :

        U_{i}=\frac{1}{2}CV^{2}

        Q=CV, is the charge on the capacitor.

Now, when the capacitor is connected to the other Capacitor, they both get some charge in equilibrium.

Hence, the total charge remains the same, the charge on each capacitor is :

        Q_{f}=\frac{Q}{2}

   Now, the energy stored is -

        U_{f}=\frac{1}{2}\frac{Q^{2}}{C}+\frac{1}{2}\frac{Q^{2}}{C}

        U_{f}=\frac{1}{4}\frac{Q^{2}}{C}

        U_{f}=\frac{1}{4}CV^{2}

Hence, the Ratio is 

        \frac{U_{i}}{U_{f}}=\frac{\frac{1}{2}CV^{2}}{\frac{1}{4}CV^{2}}=\frac{\frac{1}{2}}{\frac{1}{4}}

        \frac{U_{i}}{U_{f}}=2

Posted by

Safeer PP

View full answer

Similar Questions

Latest Question