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

Two charges $Q$ and $4Q$ are separated by a distance of $6 \text{ cm}$. The distance of the point from $4Q$ at which the net electric field is zero is: (in $\text{ cm}$)

If the potential at the centre of a uniformly charged ring is $V_0$,then the electric field at its centre will be (assume radius $= R$):

$A$ uniformly charged rod of length $4\,cm$ and linear charge density $\lambda = 30\,\mu C/m$ is placed as shown in the figure. Calculate the $x-$ component of the electric field at point $P$.

Two large circular discs separated by a distance of $0.01 \ m$ are connected to a battery via a switch as shown in the figure. Charged oil drops of density $900 \ kg \ m^{-3}$ are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity,the switch is closed to apply a voltage of $200 \ V$ across the discs. As a result,an oil drop of radius $8 \times 10^{-7} \ m$ stops moving vertically and floats between the discs. The number of electrons present in this oil drop is (neglect the buoyancy force,take acceleration due to gravity $g = 10 \ m \ s^{-2}$ and charge on an electron $e = 1.6 \times 10^{-19} \ C$):

The magnitude of electric field intensity $E$ such that an electron placed in it would experience an electrical force equal to its weight is given by: