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?

Four positive point charges $+q$ are kept at the four corners of a square of side $l$. The net electric field at the midpoint of any one side of the square is (take $\frac{1}{4 \pi \epsilon_0}=k$ ).

The linear charge densities of two infinitely long thin and parallel wires are $4 \ Cm^{-1}$ and $8 \ Cm^{-1}$,and the separation between them is $4 \ cm$. The electric field intensity at the midpoint on the line joining them 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 $x$?

The point charges $Q$ and $-2Q$ are placed at a distance $r$ apart. If the electric field at the location of $Q$ (point $A$) due to $-2Q$ is $\vec E$,then the electric field at the location of $-2Q$ (point $B$) due to $Q$ will be:

$A$ negative charge is placed at the centre of a non-conducting sphere. The direction of the electric field at any point on the surface of the sphere is