Four charges are placed at the circumference of a dial clock as shown in the figure. If the clock has only an hour hand,then the resultant force on a charge $q_0$ placed at the center points in the direction which shows the time as (in $:30$)

What is the force (in $N$) between two small charged spheres having charges of $2 \times 10^{-7} \;C$ and $3 \times 10^{-7} \;C$ placed $30 \;cm$ apart in air?

The charges on two spheres are $+7\,\mu C$ and $-5\,\mu C$ respectively. They experience a force $F$. If each of them is given an additional charge of $-2\,\mu C$,the new force of attraction will be

When the distance between the charged particles is halved,the force between them becomes

Two charges of equal magnitudes and at a distance $r$ exert a force $F$ on each other. If the charges are halved and the distance between them is doubled,then the new force acting on each charge is

$A$ charged particle of mass $m_{1}$ and charge $q_{1}$ is revolving in a circle of radius $r$. Another charged particle of charge $q_{2}$ and mass $m_{2}$ is situated at the centre of the circle. If the velocity and time period of the revolving particle are $v$ and $T$ respectively,then: