$A$ 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:

Four charges,each of magnitude $q$ coulomb,are placed at points $(-1,0,0), (1,0,0), (0,-1,0)$ and $(0,1,0)$ in the $xy$-plane. The distances along the axes are measured in metres. The magnitude of the electric field at the point $(0,0,1)$ on the $Z$-axis is

The figure shows some electric field lines corresponding to an electric field. Then:

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 $(r > R)$.

Two charges, each equal to $-q$, are kept at $(-a, 0)$ and $(a, 0)$. $A$ charge $q$ is placed at the origin. If $q$ is given a small displacement $y$ along the $y$-direction, the force acting on $q$ is proportional to:

The point charges $+q, -q, -q, +q, +Q$ and $-q$ are placed at the vertices of a regular hexagon $ABCDEF$ as shown in the figure. The electric field at the centre of the hexagon '$O$' due to the five charges at $A, B, C, D$ and $F$ is thrice the electric field at the centre '$O$' due to charge $+Q$ at $E$ alone. The value of $Q$ is: