Two point charges $q_1$ and $q_2$ are '$l$' distance apart. If one of the charges is doubled and the distance between them is halved, the magnitude of the force becomes $n$ times, where $n$ is:

For the system shown in the figure,find $Q$ such that the resultant force on $q$ is zero.

Three charges $4q, Q$ and $q$ are placed at positions $0, \frac{l}{2}$ and $l$ respectively along a straight line. If the resultant force on $q$ is zero,then $Q$ is equal to

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \; \mathring{A}$ apart is,$(m_{e} \simeq 9 \times 10^{-31} \; kg, e = 1.6 \times 10^{-19} \; C)$. (Take $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}$)

$A$ 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 $\text{ cm}$)

Two small conducting balls of identical mass $20 \text{ g}$ and identical charge $10^{-10} \text{ C}$ hang from non-conducting threads of length $L = 300 \text{ cm}$. If the equilibrium separation of the balls is $x$ and $x \ll L$,then the magnitude of $x$ is (Assume $4 \pi \varepsilon_0 = \frac{1}{9 \times 10^9} \text{ F/m}$ and $g = 10 \text{ m/s}^2$):