Two point charges $+9e$ and $+e$ are placed at a distance of $16\, cm$ from each other. Where should a third charge $q$ be placed between them so that the system remains in equilibrium?

$A$ small uncharged conducting sphere is placed in contact with an identical sphere having a charge of $4 \times 10^{-8} \text{ C}$. They are then separated to a distance such that the force of repulsion between them is $9 \times 10^{-3} \text{ N}$. The distance between them is (Take $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$ in $SI$ units). (in $\text{ cm}$)

For any fixed distance,the electromagnetic force between two protons is $10^n$ times the gravitational force between them. Then $n=$

$A$ negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field of $100 \ V m^{-1}$. If the mass of the drop is $1.6 \times 10^{-3} \ g$,the number of electrons contained in the drop is:

$A$ charge $Q$ is distributed over a line of length $L$. Another point charge $q$ is placed at a distance $r$ from the centre of the line distribution. Then the force experienced by $q$ is

The electrical force between two point charges is $200 \ N$. If we increase $10 \ \%$ charge on one of the charges and decrease $10 \ \%$ charge on the other,then the electrical force between them for the same distance becomes . . . . . . . . . . (in $N$)