Question

Calculate the electric field produced by the radiation coming from a 100W bulb at a distance of 3m. Assume that the efficiency of the bulb is 2.5 and it is a point source.

Answer 1

The electric field will be 4.07 V/m

Explanation:

The bulb behaves as a point source so, it radiates light uniformly in all directions. First we will calculate area of surrounding sphere.  

A=4πr^2 = 4π(3)^2 =113m^2

Now,

Intensity = power/area = 100 × 0.025/113 = 0.022 W/m^2

In this case half of the intensity is given by electric field and other half by magnetic field so,

½ I=1/2(∈oE^(2)rms C)

Where C is the speed of light and Erms is the root mean square value of electric field. By putting values in above equation, we get,

Erms = 2.9V/m

And,

EP= √2Erms = 4.07 V/m

Answer 2

The electric field produced by the radiation is 4.07 V/m.

Explanation:

The peak electric field strength in the radiation of the wave is given as

 $E_p=\sqrt{\frac{2I}{\epsilon_0c} }$

Here I is the intensity of the wave and c is the speed of light.

Intensity,

 $I=\frac{P}{A}$

Here P is the power and A is the surface area of surrounding of sphere.

Given P = 100 W x 2.5 %  and $A = 4\pi(3m)^2$,

$I=\frac{100\times\frac{2.5}{100} }{4\pi(3)^3}$

I = 0.022 W/m^2.

substitute in above formula, we get

$E_p=\sqrt{\frac{2\times0.022W/m^2}{8.85\times10^-^1^2C^2/N.m^2\times3\times10^8m/s} }$

$E_p=4.07 V/m$

Thus, the electric field produced by the radiation is 4.07 V/m.

#Learn More: electric field strength in the radiation.

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