The displacement of a charge $Q$ in the electric field $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$ is $\vec{r} = a \hat{i} + b \hat{j}$. The work done by the electric field is:

$A$ 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 $cm/s$.

$A$ point charge of $10^{-8} \text{ C}$ is placed at the origin. The work done in moving a point charge of $2 \mu\text{C}$ from point $A(4, 4, 2) \text{ m}$ to $B(2, 2, 1) \text{ m}$ is . . . . . . $\text{J}$.

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:

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?

Two fixed charges $A$ and $B$ of $5 \mu C$ each are separated by a distance of $6 \text{ m}$. $C$ is the midpoint of the line joining $A$ and $B$. $A$ charge $Q$ of $-5 \mu C$ is shot perpendicular to the line joining $A$ and $B$ through $C$ with a kinetic energy of $0.06 \text{ J}$. The charge $Q$ comes to rest at a point $D$. The distance $CD$ is