Physics Ncert Solutions Class 12th

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

Explanation- yes, the magnetic force differ from inertial frame to frame. The magnetic force is frame dependent.

The net acceleration which comes into existing out of this is however, frame independent for inertial frames.

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

The force on a charge particle in electric field F = qE

The free electrons (negative charge) experience electrostatic force in a direction opposite to the direction of electric field.

The direction of electric field is always from higher potential to lower. Hence direction of travel of electrons is from lower potential to region of higher potential.

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.

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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

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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 .

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$\sigma$ =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

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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