A particle of mass $m$ and charge $q$ is lying at the mid-point of two stationary particles distant $2l$ and each carrying a charge $q$. If the free charged particle is displaced from its equilibrium position through distance $x\left(x<<l\right),$ then the particle will,

In the figure shown, a charge $+q$ is fixed at point $A$. Another identical charge of mass $m$ at point $B$ is given a velocity $V$ perpendicular to the line $AB$. At this moment, the radius of curvature $R$ of the resultant path at $B$ will be,

A force $F$ acts between sodium and chlorine ions of salt (sodium chloride) when put $1 \mathrm{cm}$ apart in air. The permittivity of air and dielectric constant of water are $\varepsilon_{0}$ and $k$, respectively. When a piece of salt is put in water, electrical force acting between sodium and chlorine ions $1 \mathrm{cm}$ apart is,

The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$. Half the gap between the two charges is filled by a dielectric slab of dielectric constant $=4 $. Then the new force of repulsion between those two charges becomes 

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

Two small spheres with mass $m_1$ and $m_2$ hang from massless insulating threads of length $l_1$ and $l_2.$ The two spheres carry charges $q_1$ and $q_2$, respectively. The spheres are hung such that they are on the same horizontal level and the threads are inclined to the vertical at angles ${\theta }_1$ and ${\theta }_2.$ Which of the condition is required if ${\theta }_1={\theta }_2?$