Class 12th

This is a Short Answer Type Questions as classified in NCERT Exemplar

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.

This is a Short Answer Type Questions as classified in NCERT Exemplar

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 .

This is a Short Answer Type Questions as classified in NCERT Exemplar

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

This is a Short Answer Type Questions as classified in NCERT Exemplar

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

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

This is a Short Answer Type Questions as classified in NCERT Exemplar

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

This is a Long Answer Type Questions as classified in NCERT Exemplar

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  

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as we know F= Qq/r²= 1dyne = 1esu of charge²/1cm²

1 esu of charge = M^1/2L^3/2T^—1

• Q=xC where x is dimensionless quantity

 So F= $Q+q_2=q_1$ =1 dyne = 10^-5N

Taking x= 1/3 $Q=q_1-q_2$ ⁹

After solving we get 1/4 $q_1+q_2$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- net electric field at plate γ due to two other plate

From plate 1 ,E₁= $\frac{-Q}{S\left(2{\epsilon }_0\right)}totheleft$

From plate 2 ,E₂= $\frac{q}{S\left(2{\epsilon }_0\right)}totheright$

Total electric field E= E₁ + E₂ = $\frac{q-Q}{S\left(2{\epsilon }_0\right)}$ to the left , if Q>q

electric field at o due to plate α = $\frac{-Q}{S\left(2{\epsilon }_0\right)}totheleft$

electric field at o due to plate β = $\frac{q1}{S\left(2{\epsilon }_0\right)}totheright$

electric field at o due to plate γ = $\frac{q2}{S\left(2{\epsilon }_0\right)}totheleft$

as the electric field at o is zero therefore

As there is no loss of charge on collision

Q+q= $\frac{Q+q_2}{S\left(2{\epsilon }_0\right)}=\frac{q_1}{S\left(2{\epsilon }_0\right)}$

On solving these

 q₁= (Q+q/2)= charge on plate β

q₂= (q/2)= charge on plate γ