Kinetic energy of an electron accelerated in a potential difference of $100\, V$ is

A particle has a mass $400$ times than that of the electron and charge is double than that of a electron. It is accelerated by $5\,V$ of potential difference. Initially the particle was at rest, then its final kinetic energy will be……$eV$

A point charge $q$ of mass $m$ is suspended vertically by a string of length $l$. A point dipole of dipole moment $\overrightarrow{ p }$ is now brought towards $q$ from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is $N \times( mgh )$, where $g$ is the acceleration due to gravity, then the value of $N$ is. . . . . . (Note that for three coplanar forces keeping a point mass in equilibrium, $\frac{F}{\sin \theta}$ is the same for all forces, where $F$ is any one of the forces and $\theta$ is the angle between the other two forces)

A disk of radius $R$ with uniform positive charge density $\sigma$ is placed on the $x y$ plane with its center at the origin. The Coulomb potential along the $z$-axis is

$V(z)=\frac{\sigma}{2 \epsilon_0}\left(\sqrt{R^2+z^2}-z\right)$

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 $\beta=\frac{2 c \epsilon_0}{q \sigma}$. 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.