Point charges $+8q$ and $-2q$ are placed at $x = 0$ and $x = L$ respectively. At which point on the $X$-axis will the resultant electric field intensity be zero?

$A$ 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)

What is the electric field at a distance $x = \sqrt{8} R$ from the center on the axis of a charged ring with charge $Q$ and radius $R$?

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

Calculate the electric field at the origin due to an infinite number of charges as shown in the figure.

$A$ positive charge $q$ is placed in a spherical cavity made in a positively charged sphere. The centres of the sphere and the cavity are displaced by a small distance $\vec{l}$. The force on charge $q$ is: