$A$ 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

Two positrons $(e^+)$ and two protons $(p)$ are kept on four corners of a square of side $a$ as shown in the figure. The mass of a proton is much larger than the mass of a positron. Let $q$ denote the charge on the proton as well as the positron. Then,the kinetic energies of one of the positrons and one of the protons,respectively,after a very long time will be:

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

The distance between the plates of a parallel plate capacitor is $5d$. The positively charged plate is at $x=0$ and the negatively charged plate is at $x=5d$. Two slabs,one of a conductor and the other of a dielectric,both of equal thickness $d$,are inserted between the plates as shown in the figure. The potential versus distance graph will look like:

$(a)$ Determine the electrostatic potential energy of a system consisting of two charges $7 \; \mu C$ and $-2 \; \mu C$ (with no external field) placed at $(-9 \; cm, 0, 0)$ and $(9 \; cm, 0, 0)$ respectively.
$(b)$ How much work is required to separate the two charges infinitely away from each other?
$(c)$ Suppose that the same system of charges is now placed in an external electric field $E = A(1/r^2)$; $A = 9 \times 10^5 \; C \cdot m^{-2}$. What would the electrostatic energy of the configuration be?

$n$ small drops of the same size are charged to $V$ volts each. If they coalesce to form a single large drop,then its potential will be: