Four equal positive charges are fixed at the vertices of a square of side $L$. The $Z$-axis is perpendicular to the plane of the square. The point $z = 0$ is the point where the diagonals of the square intersect. Find the plot of the electric field $E$ due to the four charges as one moves along the $Z$-axis.

$A$ long cylindrical shell carries positive surface charge $\sigma$ in the upper half and negative surface charge $-\sigma$ in the lower half. The electric field lines around the cylinder will look like the figure given in:

Answer carefully:
$(a)$ Two large conducting spheres carrying charges $Q_{1}$ and $Q_{2}$ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $Q_{1} Q_{2} / 4 \pi \varepsilon_{0} r^{2}$,where $r$ is the distance between their centres?
$(b)$ If Coulomb's law involved $1/r^{3}$ dependence (instead of $1/r^{2}$),would Gauss's law be still true?
$(c)$ $A$ small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
$(d)$ What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
$(e)$ We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
$(f)$ What meaning would you give to the capacitance of a single conductor?
$(g)$ Guess a possible reason why water has a much greater dielectric constant $(=80)$ than,say,mica $(=6)$?

The earth's surface has a negative surface charge density of $10^{-9} \; C\;m^{-2}$. The potential difference of $400 \; kV$ between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only $1800 \; A$ over the entire globe. If there were no mechanism of sustaining the atmospheric electric field,how much time (roughly) would be required to neutralize the earth's surface (in $; s$)? (This never happens in practice because there is a mechanism to replenish electric charges,namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth $= 6.37 \times 10^{6} \; m$.)

Two fixed,identical conducting plates $(\alpha)$ and $(\beta)$,each of surface area $S$,are charged to $-Q$ and $q$,respectively,where $Q > q > 0$. $A$ third identical plate $(\gamma)$,free to move,is located on the other side of the plate with charge $q$ at a distance $d$ as per the figure. The third plate is released and collides with the plate $(\beta)$. Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst $(\beta)$ and $(\gamma)$.
$(a)$ Find the electric field acting on the plate $(\gamma)$ before collision.
$(b)$ Find the charges on $(\beta)$ and $(\gamma)$ after the collision.
$(c)$ Find the velocity of the plate $(\gamma)$ after the collision and at a distance $d$ from the plate $(\beta)$.

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: