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

$ABC$ is a right-angled triangle in which $AB = 3\,cm$ and $BC = 4\,cm$. And $\angle ABC = \pi / 2$. The three charges $+15\,e.s.u.$,$+12\,e.s.u.$,and $-20\,e.s.u.$ are placed respectively on $A$,$B$,and $C$. The force acting on $B$ is.......$dynes$.

Four charges equal to $-Q$ are placed at the four corners of a square of side length $a$,and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is:

Two identical conducting spheres carry identical charges. If the spheres are set at a certain distance apart,they repel each other with a force $F$. $A$ third conducting sphere identical to the other two,but initially uncharged,is touched to one sphere and then to the other before being removed. The force between the original two spheres is now

Whose result is the whole of electrostatics?

Three charges $4q$,$Q$,and $q$ are placed in a straight line at positions $0$,$l/2$,and $l$ respectively. The resultant force on $q$ will be zero if $Q = $