Deutron and $\alpha - $ particle are put $1\,\mathop A\limits^o $ apart in air. Magnitude of intensity of electric field due to deutron at $\alpha - $ particle is

What will be the magnitude of electric field at point $O$ as shown in figure ? Each side of the figure is $I$ and perpendicular to each other.

If the net electric field at point $\mathrm{P}$ along $\mathrm{Y}$ axis is zero, then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5 \sqrt{x}}$, where $\mathrm{x}=$. . . . . .

A positively charged thin metal ring of radius $R$ is fixed in the $xy - $ plane with its centre at the $O$. A negatively charged particle $P$ is released from rest at the point $(0,\,0,\,{z_0})$, where ${z_0} > 0$. Then the motion of $P$ is

Five point charge each having magnitude $‘q’$ are placed at the corner of hexagon as shown in fig. Net electric field at the centre $‘O’$ is $\vec E$. To get net electric field at $‘O’$ be $6\vec E$, charge placed on the remaining sixth corner should be

Give physical meaning of electric field.