Physics Ncert Solutions Class 12th

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

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Explanation- Electric flux through a dipole is always zero, because the positive and negative charges cancel each other out

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Explanation- electric field at the axis of the ring is E=KQy/ (R²+z²)^3/2 where z is distance .

F=qE=KQqy/ (R²+z²)^3/2

When z<

F= $\frac{2\pi }{w}$ = -Kz

So force is directly proportional to – distance that is completely defined that is follows S.H.M

(b)w= $\pi \sqrt[]{\frac{m}{k}}$

T=2 $\pi \sqrt[]{\frac{m4\pi {\epsilon }_o{d}^3}{2q^2}}$ w=2 $\frac{kqQy}{R3}$

T=2 $\surd \frac{k}{m}$

Explanation- two charge -q at A and B

AB=AO+OB=2d and x= small distance perpendicular to O

When x

F=qq/4 $\frac{1}{4\pi {\epsilon }_o}\frac{x^2}{\left({10}^{-2}\right)2}$ where AP=BP=r but horizontal components gets cancel out each other and vertical components gets add .

If angle APO=O the net force on q along PO is F'= 2Fcos $*10$

= $\pi {\epsilon }_0=8.98755*{10}^{9}N{m}^2{C}^{-2}$ = $\pi r^2$

When x $\theta$

K= $\frac{2q^2}{4\pi {\epsilon }_o{r}^2}\frac{x}{r}$ ,F $\frac{2q^{2}x}{4\pi {\epsilon }_o{(d^2+x^2)}^{3/2}}$

Force on charge q is proportional to its displacement from the center O and it is directed towards O

Hence we can say that motion of charge would be simple harmonic

Where w= $\frac{2q^{2}x}{4\pi {\epsilon }_o{d}^3}=kx$   and T= $\frac{2q^2}{4\pi {\epsilon }_o{d}^3}$

T= 2 $\propto x$ = 2 $\sqrt[]{\frac{k}{m}}$ = [8π³?_o md³/q²]^1/2