physics ncert exemplar solution class 12th chapter one

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Explanation- (i)The cesium atoms, are situated at the corners of a cube and Cl atom is situated at the centre of the cube. From the given figure, we can analyse that the chlorine atom is at equal distance from all the eight comers of cube where cesium atoms are placed. Thus, due to symmetry the electric field due to all Cs atoms, on Cl atom will cancel out. Hence net electric field at the centre of cube is zero.

(ii) we know, f=qE  E= Kq/r²= Ke/r²

F= e (E)=e (ke/r²)=ke²/r²

Distance= $\sqrt[]{r^2+r^2+r^2}$ = $\sqrt[]{{0.2}^2+{0.2}^2+{0.2}^2}*$ 10^-9m

F=8.99 $*$ 10⁹ (1.6 $*$ 10^-16)²/ (0.346 $*$ 10^-9)²=1.92 $*$ 10^-9N

Explanation- (i) F= $\frac{Kq2}{r2}$ = 9 $*$ 10^{9 $*$} (34.8 $*$ 10³)²/ (10^-2)²=1.09 $*$ 10²³N

                      (ii) F= $\frac{Kq2}{r2}$ = 9 $*$ 10^{9 $*$} (34.8 $*$ 10³)²/ (100)²=1.09 $*$ 10¹⁵N

                      (iii) F= $\frac{Kq2}{r2}$ = 9 $*$ 10⁹ $*$ (34.8 $*$ 10³)²/ (10⁶)²=1.09 $*$ 10⁷N

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Explanation- 1 molar mass of Al has N_A (Avogadro number)= 6.023 $*$ 10²³

M= $\frac{NA}{M}*$ m= 6.023 $*$ 10²³/27 $*$ 0.75 = 1.6 $*$ 10²²

Magnitude of positive and negative charges in one paisa coin = Nze

As aluminium has 13 electron and 13 protons so

=  1.6 $*$ 10²² $*$ 13 $*$ 1.6 $*$ 10^-19=33.28 $*$ 10³C.

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Explanation- No field inside a hollow body because if we give charge to a hollow body whole charge is distributed outside its surface. This phenomenon is called electrostatic shielding which is used to protect things from electric field or electric shock.

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Explanation- if the total charge inside a surface is zero it does not mean that electric field is zero may it will flow from inside to outside or vice versa. But if electric field is zero then charge must be zero .

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Explanation- In any neutral atom, the number of electrons and protons are equal, and the protons and electrons are bound into an atom with distinct and independent existence. Electrostatic fields are caused by the presence of excess charges. But there can be no excess charge of an isolated conductor. So, the electrostatic fields inside a conductor is zero

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Explanation – (i) surface density=charge/area= -Q/4 $\pi R$ ₁²

                        (ii) surface density=charge/area= +Q/4 $\pi R$ ₂²