Physics

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(c) Electric potential are always less in the direction of electric field

E= - $\frac{dv}{dr}$ and W= qdv . so potential energy also decrease in the direction of electric field

This is a multiple choice answer as classified in NCERT Exemplar

(d) We know that I=  $\frac{v}{R+r}$ = $\frac{2.5}{2+0.5}$ = 1 A

The potential difference across 2 ohm = 1 (2)= 2V

As Q=CV = 4 $*$ 2= 8 μC

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Let us take a point p to be at a distance z from the center of ring

V=K $\int \frac{dq}{r}=k$ $\frac{dq}{\sqrt[]{z^2+a^2}}$  = $\frac{k}{\sqrt[]{z^2+a^2}}\int dq$ = $\frac{kQ}{\sqrt[]{z^2+a^2}}$

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

Let us take a point p to be at a distance z from the center of ring

V=K $\int \frac{dq}{r}=k$ $\frac{dq}{\sqrt[]{z^2+a^2}}$  = $\frac{k}{\sqrt[]{z^2+a^2}}\int dq$ = $\frac{kQ}{\sqrt[]{z^2+a^2}}$

 

U= w =qV= $\frac{kQ(-q)}{\sqrt[]{z^2+a^2}}$ = $\frac{-KQq}{\sqrt[]{z^2+a^2}}$

Charge would perform oscillations.

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According to the relation that E= - $\frac{dv}{dr}$ , potential always decrease in the direction of electric field . so if we go from any point to infinity that decreases the potential.

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If there is a potential then E? 0 electric field comes into existence, which is given by as E=-dV/dr
It means there will be field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside. Hence, the entire volume inside must be equipotential.

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As the electric field and force both are conservative so work done in conservative body as the body is closed is zero.

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Explanation- the system will be in stable equilibrium if the net force on the system is zero and net torque on the system is also zero.

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Explanation- magnetic moment of dipole at origin o is

M=Mk

Magnetic field induction at p due to dipole moment

B= $\frac{{\mu }_0}{4\pi }\frac{2Mcos\theta }{r^3}$ r

dS= r(rsin $\theta d\theta$ )r = (r^{2 $\theta$} sind $\theta r$ )

$\int B.ds$  = $\int \frac{{\mu }_0}{4\pi }\frac{2Mcos\theta }{r^3}$ r(r²sin $\theta d\theta r$ )

$\frac{{\mu }_0}{4\pi }\frac{M}{r}{\int }_0^{2\pi }2sin\theta cos\theta d\theta$

$\frac{{\mu }_0}{4\pi }\frac{M}{r}{\int }_0^{2\pi }sin2\theta d\theta$

$\frac{{\mu }_0}{4\pi }\frac{M}{r}\left(\frac{-cos2\theta }{2}\right)$

$\frac{{\mu }_0}{4\pi }\frac{M}{r}$ {cos4 $\pi -cos0$ }

$\frac{{\mu }_0}{4\pi }\frac{M}{r}$ (1-1)= 0