An electron with kinetic energy $K _{1}$ enters between parallel plates of a capacitor at an angle $'\alpha'$ with the plates. It leaves the plates at angle $' \beta '$ with kinetic energy $K _{2}$. Then the ratio of kinetic energies $K _{1}: K _{2}$ will be ....... .
$\frac{\sin ^{2} \beta}{\cos ^{2} \alpha}$
$\frac{\cos ^{2} \beta}{\cos ^{2} \alpha}$
$\frac{\cos \beta}{\cos \alpha}$
$\frac{\cos \beta}{\sin \alpha}$
Two identical capacitors have same capacitance $C$. One of them is charged to the potential $\mathrm{V}$ and other to the potential $2 \mathrm{~V}$. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system 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
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.
A $10\, micro-farad$ capacitor is charged to $500\, V$ and then its plates are joined together through a resistance of $10\, ohm$. The heat produced in the resistance is........$J$
How does a capacitor store energy ? And obtain the formula for the energy stored in the capacitor ?