$A$ spherical conductor of diameter $6 \ mm$ is kept in a uniform electric field of intensity $2 \times 10^7 \ N/C$. The maximum charge on the conductor is $\left[\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI units}\right]$. (in $\mu C$)

$A$ point charge of $0.009 \ \mu C$ is placed at the origin. Calculate the electric field intensity due to this point charge at the point $(\sqrt{2}, \sqrt{7}, 0)$.

In the given figure,the electric field at the center $O$ due to section $AB$ of a uniformly charged ring is $\overrightarrow{E}$. What will be the electric field at $O$ due to the remaining section $ACB$?

Two non-conducting solid spheres of radii $R$ and $2R$,having uniform volume charge densities $\rho_1$ and $\rho_2$ respectively,touch each other. The net electric field at a distance $2R$ from the centre of the smaller sphere,along the line joining the centres of the spheres,is zero. The ratio $\frac{\rho_1}{\rho_2}$ can be;
$(A) -4$ $(B) -\frac{32}{25}$ $(C) \frac{32}{25}$ $(D) 4$

$A$ thin metallic wire having a cross-sectional area of $10^{-4} \, m^2$ is used to make a ring of radius $30 \, cm$. $A$ positive charge of $2 \pi \, pC$ is uniformly distributed over the ring, while another positive charge of $30 \, pC$ is kept at the centre of the ring. The tension in the ring is . . . . . . $N$; provided that the ring does not get deformed (neglect the influence of gravity). (Given, $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, SI$ units)

Three charges $2q, -q$ and $-q$ are located at the vertices of an equilateral triangle. At the center of the triangle,