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 . . . . . . $J$.

$A$ test charge $q$ is moved in the electric field of a point charge $Q$ along two different closed paths as shown in the figure. The first path has sections along and perpendicular to the lines of the electric field. The second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?

The distance between the plates of a capacitor is $1 \ mm$. An electric field of $5 \times 10^5 \ V/m$ is produced between the two plates. What will be the change in the potential energy $(PE)$ of an electron if it is moved from one plate to the other?

$A$ particle of mass $100\, g$ and charge $2\, \mu C$ is released from a distance of $50\, cm$ from a fixed charge of $5\, \mu C$. Find the speed of the particle when its distance from the fixed charge becomes $3\, m$. Neglect any other force. (Result in $m/s$) (in $.73$)

$A$ point charge $q$ moves from point $P$ to point $S$ along the path $PQRS$ (as shown in the figure) in a uniform electric field $E$ pointing parallel to the positive direction of the $X$-axis. The coordinates of the points $P, Q, R,$ and $S$ are $(a, b, 0), (2a, 0, 0), (a, -b, 0),$ and $(0, 0, 0)$ respectively. The work done by the field in the above process is given by the expression:

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: