Two point charges $3 \times 10^{-6} \,C$ and  $8 \times 10^{-6} \, C$ repel each other by a force of  $6 \times 10^{-3} \, N$. If each of them is given an additional charge $-6 \times 10^{-6} \, C$, the force between them will be

Three charge $q$, $Q$ 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 froce on $q$ zero, the charge $Q$ must be equal to

Force between two point charges $q_1$ and $q_2$ placed in vacuum at ' $r$ ' $\mathrm{cm}$ apart is $F$. Force between them when placed in a medium having dielectric $\mathrm{K}=5$ at $\mathrm{r} / 5$ $\mathrm{cm}$ apart will be:

A cube of side $b$ has a charge $q$ at each of its vertices. The electric field due to this charge distribution at the centre of this cube will be

Point charge $q$ moves from point $P$ to point $S$ along the path $PQRS$ (figure shown) in a uniform electric field $E$ pointing coparallel to the positive direction of the $X – $axis. The coordinates of the points $P,\,Q,\,R$ and $S$ are $(a,\,b,\,0),\;(2a,\,0,\,0),\;(a,\, – b,\,0)$ and $(0,\,0,\,0)$ respectively. The work done by the field in the above process is given by the expression