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  • Assume that a lamp radiates power uniformly in all directions. What is the magnitude of electric field

Assume that a lamp radiates power\mathrm{P} uniformly in all directions. What is the magnitude of electric field strength at a distance \mathrm{r} from the lamp?
 

Option: 1

\mathrm{\frac{\mathrm{P}}{\pi \mathrm{c} \varepsilon_0 \mathrm{r}^2}}

 


Option: 2

\mathrm{\frac{\mathrm{P}}{2 \pi \mathrm{c} \varepsilon_0 \mathrm{r}^2}}
 


Option: 3

\mathrm{\sqrt{\frac{\mathrm{P}}{2 \pi \varepsilon_0 \mathrm{r}^2 \mathrm{c}}}}
 


Option: 4

\mathrm{\sqrt{\frac{\mathrm{P}}{\pi \varepsilon_0 \mathrm{cr}^2}}}


Answers (1)

best_answer

A lamp radiates light uniformly in all directions. Therefore, intensity \mathrm{ I} at a distance \mathrm{ r}from the lamp is
\mathrm{ \mathrm{I}=\frac{\text { Power }}{\text { Area }}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^2} }                       (i)

Intensity of the electromagnetic wave is

\mathrm{ \mathrm{I}=<\mathrm{u}>\mathrm{c}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{c} }                   (ii)

Equating (i) and (ii), we get

\mathrm{ \frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{c}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^2} }

\mathrm{ \mathrm{E}_0=\sqrt{\frac{2 \mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2 \mathrm{c}}}=\sqrt{\frac{\mathrm{P}}{2 \pi \varepsilon_0 \mathrm{r}^2 \mathrm{c}}} }

Hence option 3 is correct.


 

Posted by

manish

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