Two identical capacitors, have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
$\frac{1}{4}C\left( {V_1^2 - V_2^2} \right)$
$\frac{1}{4}C\left( {V_1^2 + V_2^2} \right)$
$\frac{1}{4}C{\left( {{V_1} - {V_2}} \right)^2}$
$\frac{1}{4}C{\left( {{V_1} + {V_2}} \right)^2}$
Work done by an external agent in separating the parallel plate capacitor is
Two positively charged particles $X$ and $Y$ are initially far away from each other and at rest. $X$ begins to move towards $Y$ with some initial velocity. The total momentum and energy of the system are $p$ and $E$.
A capacitor $4\,\mu F$ charged to $50\, V$ is connected to another capacitor of $2\,\mu F$ charged to $100 \,V$ with plates of like charges connected together. The total energy before and after connection in multiples of $({10^{ - 2}}\,J)$ is
A $5.0\, \mu F$ capacitor is charged to a potential difference $800\, V$ and discharged through a conductor. The energy(in $J$) given to a conductor during the discharge is
Two capacitors each of $1\,\mu F$ capacitance are connected in parallel and are then charged by $200\;volts$ $d.c.$ supply. The total energy of their charges (in $joules$) is