Two point charges of equal magnitude but opposite sign are placed at a certain distance. The force between them is $F$. If $75\%$ of the charge from one is transferred to the other,what will be the new force between them?

Assume that you have $10^{23}$ carbon atoms. All the nuclei are placed at the North Pole of the Earth and all the electrons are placed at the South Pole of the Earth. (Radius of Earth = $6400 \ km$). The force between the charges is approximately:

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 $g = 10 \, ms^{-2}$]

Three charges $-q_1$,$+q_2$,and $-q_3$ are placed as shown in the figure. The $x$-component of the force on $-q_1$ is proportional to

Force of attraction between two point charges $Q$ and $-Q$ separated by $d \, \text{m}$ is $F_e$. When these charges are placed on two identical spheres of radius $R = 0.3 \, d$ whose centres are $d \, \text{m}$ apart,the force of attraction between them is

$12$ positive charges of magnitude $q$ are placed on a circle of radius $R$ in a manner that they are equally spaced. $A$ charge $Q$ is placed at the centre. If one of the charges $q$ is removed,then the force on $Q$ is