source of sinusoidal electromagnetic waves radiates uniformly in all directions. At a distance of 10.0 m from this source, the amplitude of

Question

source of sinusoidal electromagnetic waves radiates uniformly in all directions. At a distance of 10.0 m from this source, the amplitude of the electric field is measured to be 3.50 N>C. What is the electric-field amplitude 20.0 cm from the source

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Thạch Thảo 5 months 2021-08-05T22:11:38+00:00 1 Answers 19 views 0

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    2021-08-05T22:13:26+00:00

    Answer:

    175\ \text{N/C}

    Explanation:

    E_1 = Initial electric field = 3.5 N/C

    E_2 = Final electric field

    r_1 = Initial distance = 10 m

    r_2 = Final distance = 20 cm

    Electric field is given by

    E=\sqrt{\dfrac{2P}{\pi r^2c\varepsilon_0}}

    So,

    E\propto \dfrac{1}{r}

    \dfrac{E_2}{E_1}=\dfrac{r_1}{r_2}\\\Rightarrow E_2=E_1\dfrac{r_1}{r_2}\\\Rightarrow E_2=3.5\dfrac{10}{0.2}\\\Rightarrow E_2=175\ \text{N/C}

    The electric field amplitude at the required point is 175\ \text{N/C}.

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