$A$ charged particle having some mass is resting in equilibrium at a height $H$ above the centre of a uniformly charged non-conducting horizontal ring of radius $R$. The force of gravity acts downwards. The equilibrium of the particle will be stable if:

$A$ negative charge is placed at the centre of a non-conducting sphere. The direction of the electric field at any point on the surface of the sphere is

The number of electrons to be put on a spherical conductor of radius $0.1 \ m$ to produce an electric field of $0.036 \ N/C$ just above its surface is $...... \times 10^4$.

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:

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?

Three identical point charges are placed at the vertices of an isosceles right-angled triangle as shown. Which of the numbered vectors coincides in direction with the electric field at the mid-point $M$ of the hypotenuse?