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

Answer 1

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 $175\ \text{N/C}$.