The mean free path of electrons in a metal is $4 \times 10^{-8} \;m$. The electric field which can give on an average $2 \;eV$ energy to an electron in the metal will be in units of $V / m$

An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference $V$ in vacuum. The final speed of the electron will be

A negatively charged plate has charge density of $2 \times {10^{ - 6}}\,C/{m^2}$. The initial distance of an electron which is moving toward plate, cannot strike the plate, if it is having energy of $200\,eV$

In the following diagram the work done in moving a point charge from point $P$ to point $A$, $B$ and $C$ is respectively as $W_A$, $W_B$ and $W_C$ , then

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.

Three charges $Q, +q$ and $+q$ are placed at the vertices of a right -angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of $Q$ is