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 :

Three point charges $q_1, q_2, q_3$ are placed at the vertices of a triangle if force on $q_1$ and $q_2$ are $\left( {2\hat i - \hat j} \right)\,N$ and $\left( {\hat i + 3\hat j} \right)\,N$, respeactively, then what will be force on $q_3$ ?

The value of electric permittivity of free space is

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

The radius of two metallic spheres $A$ and $B$ are ${r_1}$ and ${r_2}$ respectively $({r_1} > {r_2})$. They are connected by a thin wire and the system is given a certain charge. The charge will be greater

Two positive ions, each carrying a charge $q,$ are separated by a distance $d.$ If $F$ is the force of repulsion between the ions, the number of electrons missing from each ion will be  ($e$ being the charge on an electron)