Two positively charged particles $X$ and $Y$ are initially far away from each other and at rest. $X$ begins to move towards $Y$ with some initial velocity. The total momentum and energy of the system are $p$ and $E$.

$A$ metallic shell has a point charge '$q$' kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of force?

$A$ 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.

$A$ charge $Q$ is uniformly distributed over the volume of an insulating sphere of radius $R$. $A$ thin metal shell of radius $b$ $(b > R)$ with a charge $-Q$ is placed around the sphere. The space between the sphere and the shell is filled with air. Which of the following graphs correctly represents the electric field $E$ as a function of distance $r$ from the center?

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:

In a regular polygon of $10$ sides,each corner is at a distance $R$ from the centre. Identical charges $q$ are placed at $9$ corners. At the centre,the magnitude of the electric field is $E$ and the potential is $V$. The ratio $\frac{V}{E}$ is