Six point charges are placed at the vertices of a regular hexagon of side $a$ as shown in the figure. Three charges are $+Q$ and three are $-Q$ arranged alternately. The electric field intensity at a point on the line passing through the centre $O$ and perpendicular to the plane of the figure at a large distance $x (x \gg a)$ from $O$ is (Let $\frac{1}{4 \pi \epsilon_0} = k$):

Four charges $q, 2q, -4q$ and $2q$ are placed in order at the four corners of a square of side $b$. The net field at the centre of the square is

Two charges $+5\,\mu C$ and $+10\,\mu C$ are placed $20\,cm$ apart. The net electric field at the midpoint between the two charges 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?

Charges $Q_{1}$ and $Q_{2}$ are at points $A$ and $B$ of a right-angled triangle $OAB$ (see figure). If the resultant electric field at point $O$ is perpendicular to the hypotenuse $AB$,then $Q_{1} / Q_{2}$ is proportional to:

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 $(r > R)$.