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
$\frac{1}{4}C\left( {V_1^2 - V_2^2} \right)$
$\frac{1}{4}C\left( {V_1^2 + V_2^2} \right)$
$\frac{1}{4}C{\left( {{V_1} - {V_2}} \right)^2}$
$\frac{1}{4}C{\left( {{V_1} + {V_2}} \right)^2}$
On increasing the plate separation of a charged condenser, the energy
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
A parallel plate capacitor has a uniform electric field ' $\overrightarrow{\mathrm{E}}$ ' in the space between the plates. If the distance between the plates is ' $\mathrm{d}$ ' and the area of each plate is ' $A$ ', the energy stored in the capacitor is : $\left(\varepsilon_{0}=\right.$ permittivity of free space)
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system
A capacitor of capacitance $C$ is initially charged to a potential difference of $V$ $volt$. Now it is connected to a battery of $2V$ with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be