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

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 electric charges $+q$ and $+q$ are placed at a distance $r$. The force between them is $F$. If one charge is fixed and the other moves in a circle of radius $r$,the work done will be .......

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting,massless strings of same length. At equilibrium,the angle between the strings is $\alpha$. The spheres are now immersed in a dielectric liquid of density $\rho_l = 800 \ kg \ m^{-3}$ and dielectric constant $K = 21$. If the angle between the strings remains the same after the immersion,then
$(A)$ electric force between the spheres remains unchanged
$(B)$ electric force between the spheres reduces
$(C)$ mass density of the spheres is $840 \ kg \ m^{-3}$
$(D)$ the tension in the strings holding the spheres remains unchanged

Three point charges are placed at the corners of an equilateral triangle with sides of $10 \ cm$ as shown in the figure. The resultant force on the charge at $B$ will be:

Three charges $q$,$Q$,and $4q$ are placed in a straight line of length $l$ at points distant $0$,$\frac{l}{2}$,and $l$ respectively from one end. In order to make the net force on $q$ zero,the charge $Q$ must be equal to