Two point charges $A$ and $B$,having charges $+Q$ and $-Q$ respectively,are placed at a certain distance apart and the force acting between them is $F$. If $25 \%$ of the charge of $A$ is transferred to $B$,then the force between the charges becomes

$(a)$ Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \; cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7} \; C$? The radii of $A$ and $B$ are negligible compared to the distance of separation.
$(b)$ What is the force of repulsion if each sphere is charged double the above amount,and the distance between them is halved?

Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$-axis from the origin,at which the net electric field due to these two point charges is zero is

Write the general equation for the Coulombian force on a charge ${q_1}$ due to a system of charges ${q_1}, {q_2}, \dots, {q_n}$.

Two identical conducting spheres $A$ and $B$ having charges $+q$ and $-q$ respectively are kept at a distance $d$ apart and experience a Coulombian force $F$ between them. If $50\%$ of the charge is transferred from sphere $B$ to $A$,then the new Coulombian force between them is . . . . . . .

Three charges $+3q$,$Q$,and $+q$ are placed in a straight line of length $\ell$ at points at distances $0$,$\frac{\ell}{2}$,and $\ell$ respectively. The value of $Q$ in order to have the net force on $+q$ to be zero is $Q = xq$. The value of $x$ is: