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.
Figure represents a crystal unit of cesium chloride, CsCl. The cesium atoms, represented by open circles are situated at the corners of a cube of side 0.40 nm , whereas a Cl atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the Cl atom carries an excess electron.
(i) What is the net electric field on the Cl atom due to eight Cs atoms?
(ii) Suppose that the Cs atom at the corner $A$ is missing. What is the net force now on the Cl atom due to seven remaining Cs atoms?
(i) From the given figure, we can analyse that the chlorine atom is at the centre of the cube i.e., at equal distance from all the eight corners of cube where cesium atoms are placed. Thus, due to symmetry the forces due to all Cs tons, on Cl atom will cancel out.
Hence,
$$\begin{aligned} & E=\frac{F}{q} \text { where } F=0 \\ \therefore\quad & E=0 \end{aligned}$$
(ii) Thus, net force on Cl atom at $A$ would be,
$$F=\frac{e^2}{4 \pi \varepsilon_0 r^2}$$
where, $r=$ distance between Cl ion and Cs ion.
Applying Pythagorous theorem, we get
$$\begin{aligned} r & =\sqrt{(0.20)^2+(0.20)^2+(0.20)^2} \times 10^{-9} \mathrm{~m} \\ & =0.346 \times 10^{-9} \mathrm{~m} \end{aligned}$$
Now,
$$\begin{aligned} F & =\frac{q^2}{4 \pi \varepsilon_0 r^2}=\frac{e^2}{4 \pi \varepsilon_0 r_2} \\ & =\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{\left(0.346 \times 10^{-9}\right)^2}=1.92 \times 10^{-9} \mathrm{~N} \end{aligned}$$