Question

A point sources of electromagnetic radiation
has an average power output of 1500 W. The
maximum value of electric field at a distance of
3 m from this source in Vm1 is​

Answer 1

Answer

AnswerOpen in answr app

AnswerOpen in answr appCorrect option is

AnswerOpen in answr appCorrect option isB

AnswerOpen in answr appCorrect option isB73

AnswerOpen in answr appCorrect option isB73Intensity of EM wave is given by:-

AnswerOpen in answr appCorrect option isB73Intensity of EM wave is given by:-I=4πR2P=Vav.c=21ε0E02×c⇒E0=2πR2ε0cP=2×3.14×(3)2×8.85×10−12×3×1081500=73mV

PLEASE MARK AS BRAINLIEST IF CORRECT

Answer 2

Answer:

Intensity of EM wave is given by:-

I=

4πR

2

P

=V

av

.c=

2

1

ε

0

E

0

2

×c

⇒E

0

=

2πR

2

ε

0

c

P

=

2×3.14×(3)

2

×8.85×10

−12

×3×10

8

1500

=73

m

V