
The electric field in an electromagnetic wave is given by
$E=\left( 50N{{C}^{-1}} \right)\sin \omega \left( t-\dfrac{x}{c} \right)$
Find the energy contained in a cylinder of cross-section $10c{{m}^{2}}$ and length $50cm$ along the x-axis.
Answer
537.3k+ views
Hint: You could firstly read the question properly and thus note down the given quantities. Using the value of peak electric field in the expression for electric field one could find the energy density. Then you could find the volume and take the product with energy density to find the energy contained in that volume.
Formula used:
Energy density,
$U=\dfrac{1}{2}{{\varepsilon }_{0}}{{E}_{0}}^{2}$
Complete answer:
In the question, we are given the expression for the electric field of an electromagnetic wave. From that expression, we will get the peak value of electric field as,
${{E}_{0}}=50N{{C}^{-1}}$
Now we have the energy density of the electromagnetic wave of peak electric field ${{E}_{0}}$ given by the expression,
$U=\dfrac{1}{2}{{\varepsilon }_{0}}{{E}_{0}}^{2}$
Where, ${{\varepsilon }_{0}}$ is the permittivity of the free space.
$U'=\dfrac{1}{2}\times 8.854\times {{10}^{-12}}\times {{50}^{2}}=1.1\times {{10}^{-8}}J/{{m}^{3}}$
But this would be the energy contained per unit volume, we are supposed to find the total energy contained within the whole cylinder.
We are given the cross-sectional area and length as $10c{{m}^{2}}$ and length $50cm$
Now, we could proceed with finding the volume of the cylinder which would be the product of the above quantities. Therefore,
$V=A\times l=10c{{m}^{2}}\times 50cm=500c{{m}^{3}}$
So, the net energy contained within the cylinder would be,
$U=U'\times V=1.1\times {{10}^{-8}}J{{m}^{-3}}\times 500\times {{10}^{-6}}{{m}^{3}}$
$\therefore U=5.5\times {{10}^{-12}}J$
Hence, we found the net energy stored in the whole cylinder to be $U=5.5\times {{10}^{-12}}J$
Note:
Unit plays a significant role in numerical problems such as these. You will have to either make sure that all the quantities are given in their SI units at the very beginning while noting down the values. Otherwise, you could change the unit at the very end as per the options or as the situation demands.
Formula used:
Energy density,
$U=\dfrac{1}{2}{{\varepsilon }_{0}}{{E}_{0}}^{2}$
Complete answer:
In the question, we are given the expression for the electric field of an electromagnetic wave. From that expression, we will get the peak value of electric field as,
${{E}_{0}}=50N{{C}^{-1}}$
Now we have the energy density of the electromagnetic wave of peak electric field ${{E}_{0}}$ given by the expression,
$U=\dfrac{1}{2}{{\varepsilon }_{0}}{{E}_{0}}^{2}$
Where, ${{\varepsilon }_{0}}$ is the permittivity of the free space.
$U'=\dfrac{1}{2}\times 8.854\times {{10}^{-12}}\times {{50}^{2}}=1.1\times {{10}^{-8}}J/{{m}^{3}}$
But this would be the energy contained per unit volume, we are supposed to find the total energy contained within the whole cylinder.
We are given the cross-sectional area and length as $10c{{m}^{2}}$ and length $50cm$
Now, we could proceed with finding the volume of the cylinder which would be the product of the above quantities. Therefore,
$V=A\times l=10c{{m}^{2}}\times 50cm=500c{{m}^{3}}$
So, the net energy contained within the cylinder would be,
$U=U'\times V=1.1\times {{10}^{-8}}J{{m}^{-3}}\times 500\times {{10}^{-6}}{{m}^{3}}$
$\therefore U=5.5\times {{10}^{-12}}J$
Hence, we found the net energy stored in the whole cylinder to be $U=5.5\times {{10}^{-12}}J$
Note:
Unit plays a significant role in numerical problems such as these. You will have to either make sure that all the quantities are given in their SI units at the very beginning while noting down the values. Otherwise, you could change the unit at the very end as per the options or as the situation demands.
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