Three particles of each charge $q$ are placed at the vertices of an equilateral triangle of side $L$. The work to be done to decrease the side of the triangle to $\frac{L}{2}$ is

$A$ 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?

Charges $5 \mu C$ and $10 \mu C$ are placed $1 \ m$ apart. Work done to bring these charges at a distance $0.5 \ m$ from each other is . . . . . . .
$(k = 9 \times 10^9 \ SI)$

$A$ 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:

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 $= 9 \times 10^{-31} \ kg$,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \ m^2 \ C^{-2}$]

$A$ ball of mass $1 \, g$ and charge $10^{-8} \, C$ moves from point $A$ $(V_A = 600 \, V)$ to point $B$ where the potential is zero. The velocity of the ball at point $B$ is $20 \, cm/s$. Find the velocity of the ball at point $A$ in $cm/s$.