$A$ $10\,\mu C$ charge is divided into two parts and placed at $1\,cm$ distance so that the repulsive force between them is maximum. The charges of the two parts are :

$A$ proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$-particle with velocity $v$. Initially,the $\alpha$-particle is at rest,but it is free to move. If gravity is neglected,then the minimum separation along the straight line of their motion will be:

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 $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}$)

$A$ charge $Q$ is divided into two charges $q$ and $Q-q$. The value of $q$ such that the force between them is maximum,is

Two small spheres each of mass $10 \, mg$ are suspended from a point by threads $0.5 \, m$ long. They are equally charged and repel each other to a distance of $0.20 \, m$. The charge on each of the sphere is $\frac{a}{21} \times 10^{-8} \, C$. The value of $a$ will be ...... . [Given $g = 10 \, ms^{-2}$]

Two charges of magnitude $5\, nC$ and $-2\, nC$ are placed at points $(2\, cm, 0, 0)$ and $(x\, cm, 0, 0)$ respectively in a region of space where there is no other external field. If the electrostatic potential energy of the system is $-0.5\,\mu J$,find the value of $x$ in $cm$.