$A$ deuteron and an $\alpha$-particle are separated by a distance of $1\,\mathring{A}$ in air. The magnitude of the electric field due to the deuteron at the position of the $\alpha$-particle is:

Equal charges $q$ are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side $a$. The magnitude of the electric field at point $C$ is:

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \ N/C$ normally. $A$ charged particle of mass $2 \ g$ is suspended by a silk thread of length $20 \ cm$ and remains at a distance of $10 \ cm$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{x}} \ \mu C$ where $x=$ . . . . . . . (Use $g=10 \ m/s^2$)

Two point charges $-q$ and $+q/2$ are situated at the origin $(0, 0, 0)$ and at the point $(a, 0, 0)$ respectively. The point along the $X$-axis where the electric field vanishes is:

Six point charges are kept $60^{\circ}$ apart from each other on the circumference of a circle of radius $R$ as shown in the figure. The net electric field at the centre of the circle is . . . . . . . ($ \epsilon_{0} $ is the permittivity of free space)

The electric field due to a point charge $2q$ at a distance $r$ is $E$. Now,if charge $q$ is uniformly distributed over a thin spherical shell of radius $R$,the electric field at a distance $\frac{r}{2}$ $(r \gg R)$ from the center of the thin spherical shell is $E'=$ . . . . . . .