Answer:
The peak electric field is [tex]E_o = 3593.6 V/m[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 1.2 \ kW = 1.2 *10^{3} \ W[/tex]
The cross-sectional area is [tex]A = 700 \ cm^2 = 700 *10^{-4} \ m^2[/tex]
Generally the average intensity of the microwave is mathematically represented as
[tex]I = \frac{c * \epsilon _o * E_o^2 }{2}[/tex]
Where [tex]c[/tex] is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
and [tex]\epsilon_o[/tex] is the permitivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
also [tex]E_o[/tex] is the peak electric field.
Now making [tex]E_o[/tex] the subject [tex]E_o = \sqrt{\frac{2 * I }{ c * \epsilon _o } }[/tex]
But this intensity of the microwave can also be represented mathematically as
[tex]I = \frac{ P }{A }[/tex]
substituting values
[tex]I = \frac{ 1.2 *10^{3} }{700 *10^{-4} }[/tex]]
[tex]I = 17142.85 \ W/m^2[/tex]
So
[tex]E_o = \sqrt{\frac{2 * 17142.85 }{ 3.0*10^{8}] * 8.85*10^{-12} } }[/tex]
[tex]E_o = 3593.6 V/m[/tex]
The peak electric field of the microwave is 3,593.1 V/m.
The given parameters;
The intensity of the wave is calculated as follows;
[tex]I = \frac{P}{A} \\\\I = \frac{1,200}{700 \times 10^{-4}} \\\\I = 17,142.86 \ W/m^2[/tex]
The peak electric field is calculated as follows;
[tex]E_o = \sqrt{\frac{2I}{c \varepsilon _o} } \\\\E_o = \sqrt{\frac{2\times 17,142.86}{3\times 10^8 \times 8.85 \times 10^{-12}} } \\\\E_o = 3,593.1 \ V/m[/tex]
Thus, the peak electric field of the microwave is 3,593.1 V/m.
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