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
$\frac{1}{4}C(V_1^2 - V_2^2)$
$\frac{1}{4}C(V_1^2 + V_2^2)$
$\frac{1}{4}C{\left( {{V_1} - {V_2}} \right)^2}$
$\frac{1}{4}C{\left( {{V_1} + {V_2}} \right)^2}$
Change $Q$ on a capacitor varies with voltage $V$ as shown in the figure, where $Q$ is taken along the $X$-axis and $V$ along the $Y$-axis. The area of triangle $OAB$ represents
$27$ similar drops of mercury are maintained at $10 \,V$ each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ....... times that of a smaller drop.
In a uniform electric field, a cube of side $1\ cm$ is placed. The total energy stored in the cube is $8.85\ \mu J.$ The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is.
The energy stored in a condenser of capacity $C$ which has been raised to a potential $V$ is given by
A capacitor of capacitance $6\,\mu \,F$ is charged upto $100$ $volt$. The energy stored in the capacitor is........$Joule$