$A$ metallic ring is uniformly charged as shown in the figure. $AC$ and $BD$ are two mutually perpendicular diameters. The magnitude of the electric field at $O$ due to arc $AB$ is $E$. What would be the magnitude of the electric field at $O$ due to arc $ABC$?

$A$ 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:

An infinite number of electric charges, each equal to $5 \text{ nC}$ (magnitude), are placed along the $X$-axis at $x = 1 \text{ cm}, x = 2 \text{ cm}, x = 4 \text{ cm}, x = 8 \text{ cm}, \dots$ and so on. In this setup, if the consecutive charges have opposite signs, then the electric field in $\text{N/C}$ at $x = 0$ is: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$

Two identical point charges are placed at a separation of $d$. $P$ is a point on the line joining the charges, at a distance $x$ from any one charge. The field at $P$ is $E$. $E$ is plotted against $x$ for values of $x$ from close to zero to slightly less than $d$. Which of the following represents the resulting curve?

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:

$A$ point charge of $10\,\mu C$ is placed at the origin. At what location on the $X$-axis should a point charge of $40\,\mu C$ be placed so that the net electric field is zero at $x = 2\,cm$ on the $X$-axis?