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

Three point charges $1 \ C$,$-2 \ C$,and $-2 \ C$ are placed at the vertices of an equilateral triangle of side $1 \ m$. The work done by an external force to increase the separation of the charges to $2 \ m$ in joules is ($\varepsilon_0 =$ permittivity of air).

Two capacitors of $3 \mu F$ and $6 \mu F$ are connected in series and a potential difference of $900 \ V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the combination is

$A$ capacitor,$C_1 = 6 \ \mu F$ is charged to a potential difference of $V_0 = 5 \ V$ using a $5 \ V$ battery. The battery is removed and another capacitor,$C_2 = 12 \ \mu F$ is inserted in place of the battery. When the switch $S$ is closed,the charge flows between the capacitors for some time until an equilibrium condition is reached. What are the charges ($q_1$ and $q_2$) on the capacitors $C_1$ and $C_2$ when the equilibrium condition is reached?

In the following diagram,the charge and potential difference across the $8\,\mu F$ capacitor will be nearly:

In the circuit below,$C_1 = 20 \, \mu F$,$C_2 = 40 \, \mu F$,and $C_3 = 50 \, \mu F$. If no capacitor can sustain more than $50 \, V$,then the maximum potential difference between $X$ and $Y$ is .......... $V$.