Physics

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

Surface charge density is inversely proportional to radius of the sphere so larger sphere has less charge density and vice versa.

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 $\theta d\theta$

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