# Total energy within a specific region.

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1. Feb 28, 2017

### jlmccart03

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

2. Relevant equations
E=E0(x/x0)
∫u * dV

3. 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?

2. Feb 28, 2017

### Staff: Mentor

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.

3. Feb 28, 2017

### jlmccart03

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?

4. Feb 28, 2017

### Staff: Mentor

What is the "everything else" that is u?

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

5. Feb 28, 2017

### jlmccart03

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

Last edited: Feb 28, 2017
6. Feb 28, 2017

### Staff: Mentor

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

7. Feb 28, 2017

### jlmccart03

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

8. Feb 28, 2017

### Staff: Mentor

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

9. Feb 28, 2017

### jlmccart03

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.

10. Feb 28, 2017

### Staff: Mentor

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/xo) term gets squared too.

11. Feb 28, 2017

### jlmccart03

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?

12. Feb 28, 2017

### Staff: Mentor

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

13. Feb 28, 2017

### jlmccart03

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