The distance between two charges $25\,\mu C$ and $36\,\mu C$ is $11\,cm$. At what point on the line joining the two charges will the electric field intensity be zero?

Two conducting spheres of radii $r_1$ and $r_2$ are charged to the same surface charge density. The ratio of electric fields near their surfaces is

$A$ negative charge $-q$ is placed at the midpoint between two fixed equal positive charges $Q$, separated by a distance $2d$. If the negative charge is given a small displacement $x$ $(x \ll d)$ perpendicular to the line joining the positive charges, how will the net force $(F)$ acting on it approximately depend on $x$?

$A$ charged particle having some mass is resting in equilibrium at a height $H$ above the centre of a uniformly charged non-conducting horizontal ring of radius $R$. The force of gravity acts downwards. The equilibrium of the particle will be stable if:

Two electric charges $9e$ and $3e$ are placed at a distance $r$ from each other. The point where the electric field intensity is zero is at a distance of .......

$A$ thin infinite sheet charge and an infinite line charge of respective charge densities $+\sigma$ and $+\lambda$ are placed parallel at $5 \ m$ distance from each other. Points $P$ and $Q$ are at $\frac{3}{\pi} \ m$ and $\frac{4}{\pi} \ m$ perpendicular distance from the line charge towards the sheet charge,respectively. $E_P$ and $E_Q$ are the magnitudes of the resultant electric field intensities at points $P$ and $Q$,respectively. If $\frac{E_P}{E_Q} = \frac{4}{a}$ for $2|\sigma| = |\lambda|$,then the value of $a$ is ...........