$A$ 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
$(\text{Take} \, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \, \text{SI unit})$

$A$ point charge of $0.009 \ \mu C$ is placed at the origin. Calculate the electric field intensity due to this point charge at the point $(\sqrt{2}, \sqrt{7}, 0)$.

In a region,the electric field varies as $E = 2x^2 - 4$,where $x$ is the distance in $m$ from the origin along the $x$-axis. $A$ positive charge of $1 \,\mu C$ is released with minimum velocity from infinity to cross the origin. Then:

$A$ cube of side $L$ has point charges $+q$ located at its seven vertices and $-q$ at remaining one vertex. The electric field at its centre is found to be $|E|=\alpha\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)$. The magnitude of constant $\alpha$ is

$A$ deuteron and an $\alpha$-particle are separated by a distance of $1\,\mathring{A}$ in air. The magnitude of the electric field due to the deuteron at the position of the $\alpha$-particle is: