Two point charges of $5 \ \mu C$ and $10 \ \mu C$ are separated by a distance of $1 \ m$. The work done to bring them to a distance of $0.5 \ m$ is:

Obtain the equation for the electric potential energy of a system of two electric charges in an external electric field.

Charges $-q, q, q$ are placed at the vertices $A, B, C$ respectively of an equilateral triangle of side $a$ as shown in the figure. If charge $-q$ is released keeping the remaining two charges fixed,then the kinetic energy of charge $-q$ at the instant when it passes through the midpoint $M$ of side $BC$ is:

Two point charges of $12\ \mu C$ and $8\ \mu C$ are placed $10\ cm$ apart. The work done in bringing them $4\ cm$ closer is:

An electron is released from a distance of $4 \ m$ from a stationary point charge $20 \ nC$. What will be the speed of the electron when it is $2 \ m$ away from the point charge?
[Charge of electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \ m^2 \ C^{-2}$]

Two charges $7 \ \mu C$ and $-2 \ \mu C$ are placed at $(-9, 0, 0) \ cm$ and $(9, 0, 0) \ cm$ respectively in an external field $E = \frac{A}{r^2} \hat{r}$, where $A = 9 \times 10^5 \ N/C \cdot m^2$. Considering the potential at infinity is $0$, the electrostatic energy of the configuration is . . . . . . $J$.