$A$ charge $Q$ of mass $m$ revolves around a charge $q$ due to the electrostatic attraction between them. The time period of its motion can be given by the formula:

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:

$A$ disk of radius $R$ with uniform positive charge density $\sigma$ is placed on the $xy$ plane with its center at the origin. The Coulomb potential along the $z$-axis is $V(z) = \frac{\sigma}{2\epsilon_0} (\sqrt{R^2+z^2} - z)$. $A$ particle of positive charge $q$ is placed initially at rest at a point on the $z$-axis with $z=z_0$ and $z_0 > 0$. In addition to the Coulomb force,the particle experiences a vertical force $\vec{F} = -c\hat{k}$ with $c > 0$. Let $\beta = \frac{2c\epsilon_0}{q\sigma}$. Which of the following statement$(s)$ is(are) correct?
$(A)$ For $\beta = \frac{1}{4}$ and $z_0 = \frac{25}{7}R$,the particle reaches the origin.
$(B)$ For $\beta = \frac{1}{4}$ and $z_0 = \frac{3}{7}R$,the particle reaches the origin.
$(C)$ For $\beta = \frac{1}{4}$ and $z_0 = \frac{R}{\sqrt{3}}$,the particle returns back to $z=z_0$.
$(D)$ For $\beta > 1$ and $z_0 > 0$,the particle always reaches the origin.

Two identical particles of mass $m$ carry a charge $Q$ each. Initially, one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first particle from a large distance with speed $v$. The closest distance of approach is:

An uncharged metal sphere is placed between two charged parallel plates. The pattern of the electric field lines will be:

Three identical charges are placed at the three corners of an equilateral triangle as shown in the figure. Which of the following statements is true for the electric field $E$ and electric potential $V$ at the center $O$?