Class 12th

This is a long answer type question as classified in NCERT Exemplar

Suppose the charge is slightly shifted to left so net potential energy will be

U= $\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\{\frac{-{\mathit{q}}^2}{\left(\mathit{d}-\mathit{x}\right)}+\frac{-{\mathit{q}}^2}{(\mathit{d}+\mathit{x})}\}$

U= $\frac{-{\mathit{q}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_{\mathit{o}}}\frac{2\mathit{d}}{({\mathit{d}}^2-{\mathit{x}}^2)}$

$\frac{\mathit{d}\mathit{U}}{\mathit{d}\mathit{x}}$ = $\frac{-{\mathit{q}}^{2}2\mathit{d}}{4\mathit{\pi }{\mathit{\epsilon }}_{\mathit{o}}}\frac{2\mathit{x}}{{({\mathit{d}}^2-{\mathit{x}}^2)}^2}$

System will be in equilibrium if this energy is zero

But when we double differentiate it

But when x<0 this value is negative so less than 0

This shows an unstable equilibrium

It follows the parabolic graph

This is a long answer type question as classified in NCERT Exemplar

Let any three point x,y,z  seprated by a distance 2d

Then potential due to two charges

$\frac{q1}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z-d)}^2}}+\frac{q2}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z+d)}^2}}$ =0

$\frac{q1}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z-d)}^2}}$ = $\frac{q2}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z+d)}^2}}$

X²+Y²+z²+ $\frac{\left(\frac{q1}{q2}\right)+1}{\left(\frac{q1}{q2}\right)-1}$ (2zd)+d²-0

this is equation of sphere .

This is a long answer type question as classified in NCERT Exemplar

$\sigma$ =charge/Area

Q= $\sigma$ (2 $\pi rdr$ )

V=KdQ/ $\surd (r$ ²+x²)=K $\sigma$ (2 $\pi rdr$ )/ $\surd (r$ ²+x²)

After integration we got

V= $k\frac{Q}{2\pi a}$ ₂ ( $\sqrt[]{x^2+a^2}$ )-x

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

Explanation- dW=F.dl=0

As dl = vdt

dW= Fvdt

dW= f.v=0

This is a long answer type question as classified in NCERT Exemplar

When initially K₁ is closed and K₂ is open, the capacitors C₁ and C₂ connected in series with battery

Q=CV=C (C₁C₂/C₁+C₂) = 6 $*$ 3/6+3=18 $\mu$ C

Now K₁ was opened and K₂ was closed, the battery and capacitor C, are disconnected from the circuit .

Here charge being shared by both capacitor so a common potential will develop.

V=C₁V₁+C₂V₂/C₁+C₂ = 18/3+3=3V

Q₂=3CV=3 $*$ 3=9μC

Q₃=3CV=3 $*$ 3=9μC

This is a long answer type question as classified in NCERT Exemplar

As we know that,   $\sigma$ =charge/Area

Q₁= $\sigma$ (4 $\pi r$ ²)

Q₂= $\sigma$ (4 $\pi \left(2r\right)$ ²)= 4Q₁

when they come in contact with each other potential being same and charge will be conserved.

So net charge is 5Q₁= 5 ( $\sigma$ (4 $\pi r$ ²)

Also V₁=V₂

KQ₁/R=KQ₂/2R  so Q₁=Q₂/2

Q₁= $\frac{5}{3}(\sigma$ (4 $\pi r$ ²) and Q₂= $\frac{10}{3}(\sigma$ (4 $\pi r$ ²)

$\sigma$ =5 $\sigma$ /3  and $\sigma$  =5 $\sigma$ /6

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

Explanation- i_G (G) = (I₁-I_G) (S₁+S₂+S₃) for I₁= 10mA

i_G (G+S₁) = (I₂-I_G) (S₂+S₃) for I₂= 100mA

i_G (G+S₁+S₂) = (I₃-I_G) (S₃) for I₃= 1A

S₁= 1W, S₂= 0,1W and S₃= 0.01W

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

Explanation- B (z) points in the same direction on z-axis and hence, J (L) isa monotonically function of L so B.dl = B.dl as cos 0 = 1

(b) $\int B.dl={\mu }_{o}I$ but when L $\to$ $\infty$

So B $\propto$ 1/r³

(c) the magnetic field due to a circular current carrying loop of radius in the xy plane with centre at origin at any point lying at a distance of from origin.

B= $\frac{{\mu }_{0}I{R}^2}{2{(Z^2+R^2)}^{3/2}}$

Z=Rtan $\theta$

dz=Rsec² $\theta d\theta$

${\int }_{-\infty }^{\infty }{B}_{z}dz=\frac{{\mu }_{0}I}{2}{\int }_{-\pi /2}^{\pi /2}cos\theta d\theta$ = ${\mu }_{0}I$