At a certain distance from a point charge,the electric field is $500 \ V/m$ and the potential is $3000 \ V$. What is this distance (in $m$)?

The magnitude of a point charge due to which the electric field $30 \ cm$ away has a magnitude of $2 \ N \ C^{-1}$ is:

Two point charges of magnitudes $-8 \mu C$ and $+32 \mu C$ are separated by a distance of $15 \ cm$ in air. The position of the point from the $-8 \mu C$ charge at which the resultant electric field becomes zero is (in $cm$):

$A$ thin non-conducting ring of radius $r$ has a linear charge density $\lambda = \lambda_0 \cos \phi$,where $\lambda_0$ is a constant and $\phi$ is the azimuthal angle. The magnitude of the electric field strength at the centre of the ring is

Two point charges of $20\,\mu C$ and $80\,\mu C$ are $10\,cm$ apart. Where will the electric field strength be zero on the line joining the charges from the $20\,\mu C$ charge? (in meters)

The point charges $+q, -q, -q, +q, +Q$ and $-q$ are placed at the vertices of a regular hexagon $ABCDEF$ as shown in the figure. The electric field at the centre of the hexagon '$O$' due to the five charges at $A, B, C, D$ and $F$ is twice the electric field at centre '$O$' due to charge $+Q$ at $E$ alone. The value of $Q$ is