$A$ negatively charged plate has a surface charge density of $2 \times 10^{-6} \, C/m^2$. Find the minimum initial distance of an electron moving toward the plate such that it does not strike the plate,given that its initial kinetic energy is $200 \, eV$.

$A$ 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 point charges $Q$ and $-Q/4$ are separated by a distance $x$. Then:

$A$ clock dial has point charges $-q, -2q, -3q, \ldots, -12q$ at the positions of the corresponding numbers on the dial respectively. The time at which the hour hand points in the direction of the net electric field at the centre of the dial is (Assume clock hands do not influence the net electric field). (in $:30$)

The dimension of $\frac{1}{2} \varepsilon_0 E^2$ (where $\varepsilon_0$ is the permittivity of free space and $E$ is the electric field) is:

The electric field is in the direction of the $x$-axis. The work done to move a charge of $0.2 \ C$ by a distance of $2 \ m$ at an angle of $60^\circ$ with the $x$-axis is $4 \ J$. What is the value of the electric field $E$ in $N/C$?