Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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

Suppose the distance between the plates is d and applied voltage is V_t= V₀2

Then electric field is E= $\frac{V}{d}$ sin(2 $\pi vt$ )

J_c= $\frac{1}{\rho }\frac{V_0}{d}sin\left(2\pi vt\right)$

  =  $J_0^{c}sin2\pi vt$

$J_0^c$   = $\frac{V_0}{\rho d}$

J_d= $\epsilon \frac{dE}{dt}$ = $\frac{V_0}{\rho d}$

   = $\epsilon \frac{2\pi v{V}_0}{d}$ cos(2 $\pi vt$ )

   = J₀^d cos 2 $\pi vt$

J₀^d= $\frac{2\pi V_{\epsilon }{V}_0}{d}$

$\frac{J_0^d}{J_0^c}$ = 2 $\pi V_{\epsilon }\rho$ = 2 $\pi 80{\epsilon }_0$ v $*$ 0.25 = 4 $\pi {\epsilon }_0$ v $*10$ =4/9

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

E= $\frac{\lambda e_s}{2\pi {\epsilon }_{0}a}j$

And B= $\frac{{\mu }_{0i}}{2\pi a}$ i= $\frac{{\mu }_{0\lambda v}}{2\pi a}$ i

Then S= $\frac{1}{{\mu }_0}$ {E $*B$ }= $\frac{1}{{\mu }_0}\{\frac{{\lambda }_j}{2\pi {\epsilon }_{0}a}*\frac{{\mu }_{0i}}{2\pi a}\mathrm{\lambda }\sqrt[]{i}\}$

 = $\frac{{\lambda }^{2}v}{4{\pi }^2{\epsilon }_{0a^2}}\left(j*i\right)=-\frac{{\lambda }^{2}v}{4{\pi }^2{\epsilon }_{0a^2}}k$