$A$ conducting sphere of radius $10 \; cm$ has an unknown charge. If the electric field $20 \; cm$ from the centre of the sphere is $1.5 \times 10^{3} \; N/C$ and points radially inward,what is the net charge (in $nC$) on the sphere?

$A$ circular wire loop of radius $1 \, cm$ carries a total charge $1 \times 10^{-6} \, C$ distributed uniformly over its length. If $0.01 \%$ of its length (circumference) is cut off, then the electric field at the centre of the loop due to the remaining wire is
$(\text{Take} \, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \, \text{SI unit})$

The maximum value of the electric field on the axis of a charged ring having charge $Q$ and radius $R$ is:

Two conducting spheres of radii $r_1$ and $r_2$ are charged to the same surface charge density. The ratio of electric fields near their surfaces is

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: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$

$A$ 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)