The magnitude of the average electric field normally present in the atmosphere just above the surface of the…

1. $-680 \mathrm{kC}$
2. $+ 670 kC$
3. $+ 680 \mathrm{kC}$
4. $- 670 \mathrm{kC}$

Solution

$\mathrm{E}=150 \mathrm{N}/\mathrm{C}$

We know that  $$$\varphi =\stackrel{}{}\overrightarrow{\text{E}}\stackrel{}{}·\stackrel{}{}\overrightarrow{\text{A}}$

                         $=\text{E}\times 4\pi {}^{}{\text{r}}^2$

                         $=150\times 4\pi \times {\left({}_{}\right)}^{}{\left({\text{R}}_{\text{e}}\right)}^{\left({}_{}\right)}$

Now, $$$\varphi =\frac{{}_{}}{}\frac{{\text{Q}}_{\text{in}}}{{}_{}}$





${\text{Q}}_{\text{in}}{}_{}=\varphi {}_{}{\in }_{\text{o}}{}_{}=150\times 4\pi {\left({}_{}\right)}^{}{\left({\text{R}}_{\text{e}}\right)}^{\left({}_{}\right)}{}^{}\times \text{8.85}\times 1{}^{}{0}^{}{}^{}\text{C}$

     $=150\times 4\pi {\left\{\right\}}^{}{\left\{\text{6.37}\times 1{}^{}{0}^6\right\}}^{\left\{{}^{}\right\}}{}^{}\times \text{8.85}\times 1{}^{}{0}^{}$

     $=600\pi \times \text{6.37}\times \text{6.37}\times \text{8.85}C$

     $= 680\text{kC}$

$$As Electric field is inward, $\mathrm{Q}$ will be negative

Asked in: JEE Main 2014 (09 Apr Online)

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