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

$A$ 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 $x$?

If the potential at the centre of a uniformly charged hollow sphere of radius $R$ is $V$,then the electric field at a distance $r$ from the centre of the sphere is $(r > R)$.

Six charges are placed around a regular hexagon of side length $a$ as shown in the figure. Five of them have charge $q$,and the remaining one has charge $x$. The perpendicular from each charge to the nearest hexagon side passes through the center $O$ of the hexagon and is bisected by the side.
Which of the following statement$(s)$ is(are) correct in $SI$ units?
$(A)$ When $x=q$,the magnitude of the electric field at $O$ is zero.
$(B)$ When $x=-q$,the magnitude of the electric field at $O$ is $\frac{q}{6 \pi \epsilon_0 a^2}$.
$(C)$ When $x=2q$,the potential at $O$ is $\frac{7q}{4 \sqrt{3} \pi \epsilon_0 a}$.
$(D)$ When $x=-3q$,the potential at $O$ is $\frac{3q}{4 \sqrt{3} \pi \epsilon_0 a}$.

$4 \times 10^{10}$ electrons are removed from a neutral metal sphere of diameter $20 \text{ cm}$ placed in air. The magnitude of the electric field (in $\text{N C}^{-1}$) at a distance of $20 \text{ cm}$ from its centre is:

Explain the concept of an electric field and derive the expression for the electric field due to a point charge.