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

The three charges $q / 2, q$ and $q / 2$ are placed at the corners $A , B$ and $C$ of a square of side ' $a$ ' as shown in figure. The magnitude of electric field $(E)$ at the comer $D$ of the square, is

The linear charge density on upper half of a segment of ring is $\lambda$ and at lower half, it is $-\lambda$. The direction of electrical field at centre $O$ of ring is :-

A charged water drop whose radius is $0.1\,\mu m$ is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be.......$N/C$ $(g = 10\,m{s^{ - 1}})$

A charged particle of mass $5 \times {10^{ - 5}}\,kg$ is held stationary in space by placing it in an electric field of strength ${10^7}\,N{C^{ - 1}}$ directed vertically downwards. The charge on the particle is

Five charges, $\mathrm{q}$ each are placed at the corners of a regular pentagon of side $\mathrm{'a'}$ as in 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 pentagon is replaced by $n\,-$ sided regular polygon with charge $q$ at each of its corners ?