A condenser having a capacity $2.0$ micro farad is charged to $200\;volts$ and then the plates of the capacitor are connected to a resistance wire. The heat produced in joules will be
$4 \times {10^4}\,J$
$4 \times {10^{10}}\,J$
$4 \times {10^{ - 2}}\,J$
$2 \times {10^{ - 2}}\,J$
The plates of a parallel plate capacitor of capacity $50\,\mu C$ are charged to a potential of $100\;volts$ and then separated from each other so that the distance between them is doubled. How much is the energy spent in doing so
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] :-
A parallel plate capacitor carries a charge $q$. The distance between the plates is doubled by application of a force. The work done by the force is
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 series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference $4\, V.$ When another parallel combination of $n_2$ capacitors, each of value $C_2,$ is charged by a source of potential difference $V$, it has the same (total) energy stored in it, as the first combination has. The value of $C_2,$ in terms of $C_1$ is then