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.
Consider a coin of Question 20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude 34.8 kC . Suppose that these equal charges were concentrated in two point charges separated by
(i) $1 \mathrm{~cm}\left(\sim \frac{1}{2} \times\right.$ diagonal of the one paisa coin $)$
(ii) 100 m ( length of a long building)
(iii) $10^6 \mathrm{~m}$ (radius of the earth). Find the force on each such point charge in each of the three cases. What do you conclude from these results?
Here,
$$\begin{aligned} & q= \pm 34.8 \mathrm{RC}= \pm 3.48 \times 10^4 \mathrm{C} \\ & r_1=1 \mathrm{~cm}=10^{-2} \mathrm{~m}, r_2=100 \mathrm{~m}, r_3=10^6 \mathrm{~m} \text { and } \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \\ & F_1=\frac{|q|^2}{4 \pi \varepsilon_0 r_1^2}=\frac{9 \times 10^9\left(3.48 \times 10^4\right)^2}{\left(10^{-2}\right)^2}=1.09 \times 10^{23} \mathrm{~N} \\ & F_2=\frac{|q|^2}{4 \pi \varepsilon_0 r_2^2}=\frac{9 \times 10^9\left(3.48 \times 10^4\right)^2}{(100)^2}=1.09 \times 10^{15} \mathrm{~N} \\ & F_3=\frac{|q|^2}{4 \pi \varepsilon_0 r_3^2}=\frac{9 \times 10^9\left(3.48 \times 10^4\right)^2}{\left(10^6\right)^2}=1.09 \times 10^7 \mathrm{~N} \end{aligned}$$
Conclusion from this result We observe that when $\pm$ charges in ordinary neutral matter are separated as point charges, they exert an enormous force. Hence, it is very difficult to disturb electrical neutrality of matter.