JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 3)
The ratio of average electric energy density and total average energy density of electromagnetic wave is :
1
3
2
$$\frac{1}{2}$$
Explanation
Avg electric energy density $=\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2$
Total Avg energy density $=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2$
$$ \therefore $$ Ratio of average electric energy density and total Avg energy density
$$ = \frac{\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2}{\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2}=\frac{2}{4}=\frac{1}{2} $$
Total Avg energy density $=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2$
$$ \therefore $$ Ratio of average electric energy density and total Avg energy density
$$ = \frac{\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2}{\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2}=\frac{2}{4}=\frac{1}{2} $$
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