What is the Peak Electric Field 2 Metres from a 60W Light Bulb?

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SUMMARY

The peak electric field 2 metres away from a 60W light bulb is calculated using the Poynting vector, which relates the electric field (E) and magnetic field (B) in electromagnetic radiation. The correct formula for the electric field is derived from the total power emitted, leading to the conclusion that the peak electric field is approximately 1.37 microvolts (0.00000137 Vm^-1). The discussion highlights the importance of correctly applying the area of a sphere in calculations and clarifies the relationship between E and B, emphasizing that B equals c times E, not mu_0 times E.

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Homework Statement



What is the peak electric field 2 metres away from a 60W light bulb. Assume that the light emits light evenly in all directions (spherically uniform).

Homework Equations



I figure that this is to do with the poynting vector.
S = \frac{1}{\mu_0}(\vec{E} \times \vec{B})

The Attempt at a Solution



The total surface integral of the poynting vector at 2 metres away from the bulb should give 60W. From this, the E field can be worked out (Since the magnitude of B is E/mu).

i.e.

\int{\vec{S} \cdot \vec{dA}} = \int{\frac{1}{\mu_0}(\vec{E} \times \vec{B}) \cdot \vec{dA}}
and since the emission is spherically symmetric, S can be taken outside of the integral. This gives that
\frac{1}{\mu_0}(\vec{E} \times \vec{B}) \times \frac{4}{3}\pi 2^2 = 60 (r=2)

Since [STRIKE]B_0 = \frac{E_0}{\mu_0}[/STRIKE]

E_0 = \frac{B_0}{\mu_0}​
this means that E_0 = \sqrt{\frac{60 \times {\mu_0}^2}{\frac{4}{3}\pi\times 4}} = 0.0021 Vm^{-1}

Is this correct?
 
Last edited:
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No. The 60 watts isn't radiated away except as blackbody radiation. Although that counts, the exercise might be about the dipole electric field of a 120 volt generator connected to a 240 ohm resistor.

Edit: you're probably right. The problem says that the light goes in all directions, not that the filament has such and such dimensions.
 
Sorry I noticed an error (area of the sphere)
A=4 \pi r^2
 
zzzoak said:
Sorry I noticed an error (area of the sphere)
A=4 \pi r^2

Haha, that's embarrassing! I'd also forgotten to square the mu_0 in my numerical answer.

That gives that the final answer is 0.00000137 Vm^-1 or 1.37 microvolts.
 
In addition, B_0 doesn't equal mu_0 times E_0, it equals c times E_0!

But that gives me an answer of 21 Volts! I'm pretty sure that this would pose some problems in real life, so that's a ridiculous answer! Any help please??
 
yeah i think you should use
B = E/c ;)

if you write out the equations, its easier to see what you've done
 

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