# Electric field around a light bulb

1. May 30, 2010

### DeShark

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

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

3. 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: May 31, 2010
2. May 30, 2010

### Antiphon

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.

3. May 31, 2010

### zzzoak

Sorry I noticed an error (area of the sphere)
$$A=4 \pi r^2$$

4. Jun 1, 2010

### DeShark

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.

5. Jun 3, 2010

### DeShark

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??

6. Jun 3, 2010

### lanedance

yeah i think you should use
B = E/c ;)

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