$A$ ring of radius $R$ is charged uniformly with a charge $+Q$. The electric field at a point on its axis at a distance $r$ from any point on the ring will be

Two point charges $a$ and $b$,whose magnitudes are the same,are positioned at a certain distance from each other with $a$ at the origin. $A$ graph is drawn between the electric field strength $E$ at points between $a$ and $b$ and the distance $x$ from $a$. $E$ is taken as positive if it is along the line joining $a$ to $b$. From the graph,it can be decided that:

$A$ thin metallic wire having a cross-sectional area of $10^{-4} \, m^2$ is used to make a ring of radius $30 \, cm$. $A$ positive charge of $2 \pi \, pC$ is uniformly distributed over the ring, while another positive charge of $30 \, pC$ is kept at the centre of the ring. The tension in the ring is . . . . . . $N$; provided that the ring does not get deformed (neglect the influence of gravity). (Given, $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, SI$ units)

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:

$A$ 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:

Why is an electric force conservative?