The lower plate of a parallel plate capacitor is supported on a rigid rod. The upper plate is suspended from one end of a balance. The two plates are joined together by a thin wire and subsequently disconnected. The balance is then counterpoised. Now a voltage $V = 5000\, volt$ is applied between the plates. The distance between the plates is $d =5\, mm$ and the area of each plate is $A = 100 cm^2.$ Then find out the additional mass placed to maintain balance.......$g$ [All the elements other than plates are massless and nonconducting] :-
$44$
$4.4$
$0.44$
$440$
A capacitor of capacity $C$ is connected with a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential $V$ again, the energy given by the battery will be
A parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E _2$. The ratio $E _2 / E _1$ is
A $60\; pF$ capacitor is fully charged by a $20\; \mathrm{V}$ supply. It is then disconnected from the supply and is connected to another uncharged $60 \;pF$ capactior is parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $nJ$)
Energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential difference $V$ is given by
The energy stored in the condenser is