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 . . . . . . $J$.

$A$ particle of mass $1 \text{ g}$ and charge $1 \mu\text{C}$ is held at rest on a frictionless horizontal surface at a distance of $1 \text{ m}$ from a fixed charge of $2 \text{ mC}$. If the particle is released, it will be repelled. What is the speed of the particle when it is at a distance of $10 \text{ m}$ from the fixed charge (in $\text{ m s}^{-1}$)?

Explain electric potential energy. Show that the sum of kinetic energy and electric potential energy remains constant.

$A$ particle $A$ has charge $+q$ and a particle $B$ has charge $+4q$,with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference,the ratio of their speeds $\frac{v_A}{v_B}$ will be:

Given $q_1 = +2 \times 10^{-8} \ C$ and $q_2 = -0.4 \times 10^{-8} \ C$. When a charge $q_3 = 0.2 \times 10^{-8} \ C$ is moved from $C$ to $D$,what is the change in the potential energy of $q_3$?

If a charge is moved against the Coulomb force of an electric field, then: