Total energy within a specific region.

[jlmccart03]

Homework Statement

A cubical region 1.0 m on a side is located between x=0 and x=1 m. The region contains an electric field whose magnitude varies with x but is independent of y and z: E=E0(x/x0), where E0 = 32 kV/m and x0 = 6.0 m . The answer needs to be in μ J.

Find the total energy in this region.

Homework Equations

E=E0(x/x0)
∫u * dV

The Attempt at a Solution

So I took the integral and got 1/2ε₀(E₀)(x/x₀)^2. When I plug in the numbers I get 3.933 μJ which is wrong? Am I missing a step?

 

[gneill]

Exactly what integration did you set up? If you check the units on what you currently have you'll find that they don't work out to Joules.

I'd expect the result to involve the square of the field strength and involve some measure of the volume.

 

[jlmccart03]

Exactly what integration did you set up? If you check the units on what you currently have you'll find that they don't work out to Joules.

I'd expect the result to involve the square of the field strength and involve some measure of the volume.

I did use the volume as just 1² since there is no dependence in the y and z directions so it was all in the x. So I simply used the integral ∫u * dV where dV is 1² and u is everything else. The constants are pulled out leaving me with an x which becomes x² along with a 1/2 out front. I believe that should be correct?

 

[gneill]

What is the "everything else" that is u?

For an electric field the energy per unit volume (energy density) is given by $\frac{1}{2}ε_o E^2$.

 

[jlmccart03]

What is the "everything else" that is u?

For an electric field the energy per unit volume (energy density) is given by $\frac{1}{2}ε_o E^2$.

The u is simply the E=E0(x/x0) is what I did. So I had the integral of ∫ u dV = ∫ 1/2 ε₀(E₀/x₀) dV = 1/2 ε₀(E₀/x₀)²x²

 

[gneill]

What happened to the x in the (x/x0) term? When you square it it should have become x² inside the integral.

 

[jlmccart03]

What happened to the x in the (x/x0) term? When you square it it should have become x² inside the integral.

Yeah I miswrote and forgot to put it in. I edited my previous post which includes the x^2 variable.

 

[gneill]

No, the x² is within the integral, so it should integrate to an x³ term in the result.

 

[jlmccart03]

No, the x² is within the integral, so it should integrate to an x³ term in the result.

Wait, where did the x^2 come from? I only have an x in the original integral. That should become x^2 not x^3.

 

[gneill]

As I stated before, the energy density is $\frac{1}{2}ε_o E^2$. Your E is given by $E = E_o \left( \frac{x}{x_o} \right)$. You'll be integrating the energy density over the volume.

When you square E for your integral, the (x/x_o) term gets squared too.

 

[jlmccart03]

As I stated before, the energy density is $\frac{1}{2}ε_o E^2$. Your E is given by $E = E_o \left( \frac{x}{x_o} \right)$. You'll be integrating the energy density over the volume.

When you square E for your integral, the (x/x_o) term gets squared too.

OHHH, ok I didn't take that into account so does that mean you get the integral to be 1/6ε₀E₀(x³/x₀) since everything, but the x is a constant can be pulled out prior to integration. Correct?

 

[gneill]

The $E_o$ and $x_o$ should end up squared. They were part of the E that was squared inside the integral. Otherwise, that looks much better

 

[jlmccart03]

The $E_o$ and $x_o$ should end up squared. They were part of the E that was squared inside the integral. Otherwise, that looks much better

Yup, got the correct answer! Thanks for the help!