Energy density of electric field

In summary, the question asks for the magnitude of the E-field at a distance of 1m for a 40W light bulb and for sunlight on Earth's surface, assuming 10% of the energy is radiated isotropically in the form of light. The energy density equation for a magnetic field is used to convert the power output into energy per unit area or volume. The intensity for both scenarios can be found by dividing the energy by the velocity at which electromagnetic energy travels.
  • #1
Cheetox
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Homework Statement



Assume that 10% of the energy dissipated by a 40W light bulb is radiated isotropically in the form of light. What is the magnitude of he E-field at a distance of 1m? What is it for sun light on the Earth's surface, given the Sun provides ~ 1400W/m2>

Homework Equations



Energy density = 1/2[tex]\epsilon[/tex][tex]_{}0[/tex]E[tex]^{}2[/tex]

The Attempt at a Solution



Ok so I know that we have 4W available, but how do I convert this into and energy per unit area on the surface or volume? I assume we have to use the equation for the energy denstiy of a magnetic field given above.
 
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  • #2
If it spreads out isotropically in all directions, then the power [4 W] is uniform at any point on the sphere of the radius, and can be given by:

[tex]
I = \frac{1}{\pi}~Wm^{-2}
[/tex]

You have the same thing i.e. Intensity for the 'sun' question. Now, Intensity is simply:

[tex]
I = \frac{Energy \times velocity}{Volume}
[/tex]

so.. i guess I've given u enough hint now.. [at what velocity does e.m energy travel?]
 
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  • #3


To solve this problem, we first need to calculate the energy dissipated by the 40W light bulb. Since we are assuming that 10% of the energy is radiated isotropically in the form of light, we can calculate the energy dissipated as:

Energy dissipated = 40W * 0.1 = 4W

Next, we need to convert this energy into an energy per unit area on the surface. To do this, we can use the equation for energy density of an electric field:

Energy density = 1/2ε0E^2

Where ε0 is the permittivity of free space and E is the magnitude of the electric field.

Since we are looking for the magnitude of the electric field at a distance of 1m, we can rearrange the equation to solve for E:

E = √(2*Energy density/ε0)

Now, we can plug in the values we have to calculate the magnitude of the electric field:

E = √(2*4W/ε0) = 0.037 N/C

Thus, the magnitude of the electric field at a distance of 1m from the 40W light bulb is 0.037 N/C.

For sunlight on Earth's surface, we are given that the Sun provides approximately 1400W/m2. Using the same equation as above, we can calculate the magnitude of the electric field at Earth's surface as:

E = √(2*1400W/ε0) = 2.87 * 10^5 N/C

Therefore, the magnitude of the electric field from sunlight on Earth's surface is 2.87 * 10^5 N/C. This is significantly higher than the electric field from the 40W light bulb, which makes sense since the Sun provides much more energy.
 

1. What is the definition of energy density of electric field?

The energy density of electric field is the amount of energy per unit volume that is stored in an electric field. It is a measure of the strength of the electric field at a particular point in space.

2. How is energy density of electric field calculated?

The energy density of electric field is calculated by multiplying the square of the electric field strength by the permittivity of the medium in which the electric field exists.

3. What is the unit of measurement for energy density of electric field?

The unit of measurement for energy density of electric field is joules per cubic meter (J/m^3).

4. Why is energy density of electric field important?

The energy density of electric field is important because it helps to quantify the amount of energy that is stored in an electric field, which is crucial for understanding and predicting the behavior of electric fields in various applications, such as in electronics and power systems.

5. How does the energy density of electric field change in different mediums?

The energy density of electric field can vary depending on the permittivity of the medium in which the electric field exists. In materials with higher permittivity, the energy density of electric field will be higher, meaning that more energy can be stored in the electric field.

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