A) 5.48 V/m
B) 7.75 V/m
C) 1.73 V/m
D) 2.45 V/m
Correct Answer: D
Solution :
\[I=\frac{E_{0}^{2}}{2c{{\mu }_{0}}}\] \[\frac{P}{4\pi {{r}^{2}}}=\frac{E_{0}^{2}}{2c{{\mu }_{0}}}\] \[E=\frac{P2c{{\mu }_{0}}}{4\pi {{r}^{2}}}\] \[E=\frac{0.1\times 2\times 3\times {{10}^{-8}}\times 4\pi \times {{N}^{-1}}}{4\pi }\] \[=\sqrt{6}=2.45\]ion
You need to login to perform this action.
You will be redirected in
3 sec
