The dimensions of an atom are of the order of an Angstrom. Thus, there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?
The protons and electrons are bound into a atom with distinct and independent existence and neutral in charge. Electrostatic fields are caused by the presence of excess charges. But there can be no excess charge on the inter surface of an isolated conductor. So, the electrostatic fields inside a conductor is zero despite the fact that the dimensions of an atom are of the order of an Angstrom.
If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.
Gauss' law also implices that when the surface is so chosen that there are some charges inside and some outside.
The flux in such situation is given by $\oint$ E.dS $=\frac{q}{\varepsilon_0}$.
In such situations, the electric field in the LHS is due to all the charges both inside and outside the surface. The term $q$ on the right side of the equation given by Gauss' law represent only the total charge inside the surface.
Thus, despite being total charge enclosed by a surface zero, it doesn't imply that the electric field everywhere on the surface is zero, the field may be normal to the surface.
Also, conversely if the electric field everywhere on a surface is zero, it doesn't imply that net charge inside it is zero.
i.e., $$\text { Putting } E=0 \text { in } \oint \text { E.dS }=\frac{q}{\varepsilon_0}$$
we get $$q=0.$$
Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure.
Thus, the electric field lines will start from positive charges and move towards infinity as given in the figure below
What will be the total flux through the faces of the cube as given in the figure with side of length $a$ if a charge $q$ is placed at?
(a) $A$ a corner of the cube
(b) $B$ mid-point of an edge of the cube
(c) $C$ centre of a face of the cube
(d) $D$ mid-point of $B$ and $C$
(a) There are eight corners in a cube so, total charge for the cube is $\frac{q}{8}$. Thus, electric flux at $A=\frac{q}{8 \varepsilon_0}$.
(b) When the charge $q$ is place at $B$, middle point of an edge of the cube, it is being shared equally by 4 cubes. Therefore, total flux through the faces of the given cube $=q / 4 \varepsilon_0$.
(c) When the charge $q$ is placed at $C$, the centre of a face of the cube, it is being shared equally by 2 cubes. Therefore, total flux through the faces of the given cube $=q / 2 \varepsilon_0$.
(d) Similarly, when charge $q$ is placed at $Q$, the mid-point of $B$ and $C$, it is being shared equally by 2 cubes. Therefore, total flux through the faces of the given cube $=q / 2 \varepsilon_0$.
A paisa coin is made up of $\mathrm{Al}-\mathrm{Mg}$ alloy and weight 0.75 g . It has a square shape and its diagonal measures 17 mm . It is electrically neutral and contains equal amounts of positive and negative charges.
Here, given quantities are
Mass of a paisa coin $=0.75 \mathrm{~g}$ Atomic mass of aluminium $=26.9815 \mathrm{~g}$
Avogadro's number $=6.023 \times 10^{23}$
$\therefore \quad$ Number of aluminium atoms in one paisa coin,
$$N=\frac{6.023 \times 10^{23}}{26.9815} \times 0.75=1.6742 \times 10^{22}$$
As charge number of Al is 13 , each atom of Al contains 13 protons and 13 electrons.
$\therefore \quad$ Magnitude of positive and negative charges in one paisa coin $=N$ ze
$$\begin{aligned} & =1.6742 \times 10^{22} \times 13 \times 1.60 \times 10^{-19} \mathrm{C} \\ & =3.48 \times 10^4 \mathrm{C}=34.8 \mathrm{kC} \end{aligned}$$
This is a very large amount of charge. Thus, we can conclude that ordinary neutral matter contains enormous amount of $\pm$ charges.