Question

The amplitude of electric field is 0.5V/m , 5km from a radio transmitter.what is the total power emitted by a transmitter

Answer 1

Power = Intensity/4πr²

Intensity of electric field = 1/2(epsilon knot)E²

r =5km = 5x10³ m

Power = 1/2*8.85x10^(-12)*(0.5)²/4π(5*10³)²

Answer 2

Answer:

The power emitted by the transmitter is 104 kW

Explanation:

Here we have been given to find the value of the power which is emitted by a particular transmitter that has an amplitude of the electric field to be 0.5 V/m and is located 5 km from a respective radio transmitter.

Now we know that the formula for calculating power is as below;

Power (P) = Intensity (I) × 4π $r^{2}$

Intensity (I) = ε₀ × (E₀ / √2 )² × c

E₀ = 0.5 V/m

c = 3 × $10^{8} m/s$

I =  ε₀ × $(0.5 / \sqrt{2} )^{2}$ V/m × 3 × $10^{8} m/s$

⇒ I = ε₀ × 0.375 × $10^{8}$ W / $m^{2}$

r = the distance of the electric field from the transmitter

= 5 km

= 5000 m

∴ P = ε₀ × 0.375 × $10^{8}$ × 4π × $(5000)^{2}$ W

⇒ P = 4πε₀ × 0.375 × $10^{8}$ × 25 × $10^{6}$ W

⇒ P = $\frac{1}{9 * (10 ^{9}) }$× 9.375 × $10^{14}$ W

⇒ P = 1.04 × $10 ^{5}$ W

⇒ P = 104000 W

⇒ P =  104 kW

Therefore the power emitted by the transmitter is 104 kW