In the rectangle shown below,two corners have charges $q_1 = -5\,\mu C$ and $q_2 = +2.0\,\mu C$. The work done in moving a charge $q_0 = +3.0\,\mu C$ from $B$ to $A$ is ......... $J$. (Given: $1/4\pi\varepsilon_0 = 10^{10}\,N\cdot m^2/C^2$)

The work done in carrying a charge $Q$ once around a circle of radius $r$ about a charge $q$ at the centre is

A charge $+q$ is fixed at each of the points $x = x_0, x = 3x_0, x = 5x_0, \dots, \infty$ on the $x$-axis, and a charge $-q$ is fixed at each of the points $x = 2x_0, x = 4x_0, x = 6x_0, \dots, \infty$. Here $x_0$ is a positive constant. Take the electric potential at a point due to a charge $Q$ at a distance $r$ from it to be $Q/(4\pi\varepsilon_0 r)$. Then, the potential at the origin due to the above system of charges is:

Three charges $q, \sqrt{2}q$,and $2q$ are placed at the corners $A, B$,and $C$ respectively of a square $ABCD$ of side length $a$. Find the electric potential at point $D$.

$A$ particle '$A$' has charge '$+q$' and a particle '$B$' has charge '$+4q$'. Each has the same mass '$m$'. When they are allowed to fall from rest through the same potential difference '$V$',the ratio of their speeds will be (particle $A$ to particle $B$):

$A$ spherical shell of radius $10 \,cm$ is carrying a charge $q$. If the electric potential at distances $5 \,cm$, $10 \,cm$, and $15 \,cm$ from the centre of the spherical shell is $V_{1}$, $V_{2}$, and $V_{3}$ respectively, then: