1.5 વક્રીભવનાંક ધરાવતા કાચના એક ચોસલા પર કોઈ | vedclass

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Question

$I=\frac{1}{2} \varepsilon_{0} E_{0}^{2} C$    ..... $(i)$

$0.96\, \mathrm{I}=\frac{1}{2} \mathrm{E}_{0}^{2} \mathrm{V}$    ...

Vedclass · Accepted Answer

$24$

Vedclass · Answer

$I=\frac{1}{2} \varepsilon_{0} E_{0}^{2} C$    ..... $(i)$

$0.96\, \mathrm{I}=\frac{1}{2} \mathrm{E}_{0}^{2} \mathrm{V}$    ..... $(ii)$

$\Rightarrow \quad 0.96=\left(\frac{E_{0}^{\prime}}{E_{0}}ight)^{2} \frac{\varepsilon}{\varepsilon_{0}} \frac{V}{C}$

$0.96 = {\left( {\frac{{E_0^\prime }}{{{E_0}}}} ight)^2}\frac{\varepsilon }{{{\varepsilon _0}}}\frac{1}{{1.5}}$

${0.96=\left(\frac{E_{0}^{\prime}}{E_{0}}ight)^{2} \varepsilon_{r} \frac{1}{1.5}}$   ..... $(iii)$

and    ${	ext{v}} = \frac{1}{{\sqrt {{\mu _{	ext{o}}}{\varepsilon _{	ext{o}}}{\mu _{	ext{r}}}{\varepsilon _{	ext{r}}}} }};$  ${	ext{v}} = \frac{{	ext{C}}}{{\sqrt {{\mu _{	ext{r}}}{\varepsilon _{	ext{r}}}} }}$

$\sqrt {{\mu _r}{\varepsilon _r}}  = \frac{C}{v}$ ; $\sqrt {{\varepsilon _r}}  = 1.5\,;$ ${\mu _r} \approx 1$ for transparent medium.

From equation $(iii)$

$0.96=\left(\frac{E_{0}^{\prime}}{E_{0}}ight)^{2}(1.5)^{2}\left(\frac{1}{1.5}ight)$

$\Rightarrow E_{\circ}^{\prime}=24 \mathrm{V} / \mathrm{m}$

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