$A$ 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:

Five charges,each $q$,are placed at the corners of a regular pentagon of side $a$ as shown in the figure.
$(a)$ $(i)$ What will be the electric field at $O$,the centre of the pentagon?
$(ii)$ What will be the electric field at $O$ if the charge from one of the corners (say $A$) is removed?
$(iii)$ What will be the electric field at $O$ if the charge $q$ at $A$ is replaced by $-q$?
$(b)$ How would your answer to $(a)$ be affected if the pentagon is replaced by an $n$-sided regular polygon with charge $q$ at each of its corners?

$A$ uniform electric field of $500 \ Vm^{-1}$ is directed at $30^{\circ}$ with the positive $X$-axis as shown in the figure. The potential difference $(V_B - V_A)$ if $OA = 3 \ m$ and $OB = 5 \ m$ is

When a $5 C$ charge is kept in a uniform electric field,a force of $5000 N$ acts on it. Find the potential difference between two points in that field,separated by a distance of $1 cm$. (in $V$)

Four identical charges are placed at the four corners of a square. If the electric field intensity at the center of the square due to any one charge is $E$,then the resultant electric field intensity at the center of the square will be .......

The magnitude of electric field intensity $E$ such that an electron placed in it would experience an electrical force equal to its weight is given by: