If the potential of a capacitor having capacity of $6\,\mu F$ is increased from $10\, V$ to $20\, V$, then increase in its energy will be
$12 \times 10^{-6} \,J$
$9 \times 10^{-4} \,J$
$4 \times 10^{-6} \,J$
$4 \times 10^{-9} \,J$
A battery does $200 \,J$ of work in charging a capacitor. The energy stored in the capacitor is ......... $J$
A $6\,\mu F$ capacitor is charged from $10\;volts$ to $20\;volts$. Increase in energy will be
Two identical capacitors are connected in parallel across a potenial difference $V$. After they are fully charged, the positive plate of first capacitor is connected to negative plate of second and negative plate of first is connected to positive plate of other. The loss of energy will be
Effective capacitance of parallel combination of two capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is $10\; \mu \mathrm{F}$. When these capacitors are individually connected to a voltage source of $1\; \mathrm{V},$ the energy stored in the capacitor $\mathrm{C}_{2}$ is $4$ times that of $\mathrm{C}_{1}$. If these capacitors are connected in series, their effective capacitance will be
How does a capacitor store energy ? And obtain the formula for the energy stored in the capacitor ?