If an electron enters into a space between the plates of a parallel plate capacitor at an angle $\alpha $ with the plates and leaves at an angle $\beta $ to the plates, the ratio of its kinetic energy while entering the capacitor to that while leaving will be
${\left( {\cos \,\alpha /\cos \,\beta } \right)^2}$
${\left( {\cos \,\beta /\cos \,\alpha } \right)^2}$
${\left( {\sin \,\alpha /\sin \,\beta } \right)^2}$
${\left( {\sin \,\beta /\sin \,\alpha } \right)^2}$
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
A parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and area of each plate is $A,$ the energy stored in the capacitor is
A piece of cloud having area $25 \times {10^6}\,{m^2}$ and electric potential of ${10^5}$ $volts$. If the height of cloud is $0.75\,km$, then energy of electric field between earth and cloud will be.....$J$
The capacity of a condenser is $4 \times {10^{ - 6}}$ farad and its potential is $100\,\,volts$. The energy released on discharging it fully will be.......$Joule$
A $2 \ \mu F$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$ is