Four identical point masses, each of mass $m$ and carrying charge $+q$, are placed at the corners of a square of side $a$ on a frictionless horizontal surface. If the particles are released simultaneously, the kinetic energy of the system when they are infinitely far apart is

Two charges,each equal to $q$,are kept at $x = -a$ and $x = a$ on the $x$-axis. $A$ particle of mass $m$ and charge $q_0 = \frac{q}{2}$ is placed at the origin. If charge $q_0$ is given a small displacement $(y << a)$ along the $y$-axis,the net force acting on the particle is proportional to

Two identical small spheres carry charges $Q_1$ and $Q_2$ $(Q_1 >> Q_2)$. The force between them is $F_1$. The spheres are brought into contact and then placed at the same distance. The new force between them is $F_2$. Then $F_1/F_2$ will be:

$A$ particle of mass $m$ and charge $q$ is kept at the top of a fixed frictionless sphere. $A$ uniform horizontal electric field $E$ is switched on. The particle loses contact with the sphere when the line joining the center of the sphere and the particle makes an angle $45^{\circ}$ with the vertical. The ratio $\frac{qE}{mg}$ is:

An infinitely long thin wire,having a uniform charge density per unit length of $5 \text{ nC/m}$,is passing through a spherical shell of radius $1 \text{ m}$,as shown in the figure. $A$ $10 \text{ nC}$ charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static,the magnitude of the potential difference between points $P$ and $R$,in Volt,is. . . .
[Given: In $SI$ units $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9, \ln 2=0.7$. Ignore the area pierced by the wire.]

Six point charges are placed at the vertices of a regular hexagon of side $a$ as shown. If $E$ represents the electric field and $V$ represents the electric potential at the center $O$,then: