Find Energy in Volume Defined by -1<x<1,-1<y<y,-1<z<1

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Discussion Overview

The discussion revolves around finding the energy stored in a volume defined by the inequalities -1

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests using a triple integral to find the energy stored in the specified volume.
  • Another participant questions the interpretation of the potential function and suggests it might represent an electric field, prompting a discussion about energy density.
  • Participants clarify that the potential V stands for volts, and the energy density of an electric field is given by (1/2)(epsilon)E^2.
  • There is a discussion about deriving the electric field from the potential, with some suggesting to take partial derivatives with respect to x, y, and z.
  • One participant calculates the gradient of the potential and discusses the implications of integrating it, expressing uncertainty about the legality of using the magnitude with variables still present.
  • Another participant emphasizes the need to square the magnitude of the electric field when calculating energy density.
  • There are calculations presented for the electric field and its magnitude, leading to discussions about integration results and potential mistakes in earlier calculations.
  • One participant raises a concern about the sign of the energy result, suggesting it should be negative, while another clarifies that energy density is always positive regardless of direction.

Areas of Agreement / Disagreement

Participants express various interpretations of the problem and calculations, leading to some disagreements about the correct approach and results. There is no consensus on the final answer or the implications of the calculations.

Contextual Notes

Participants note missing units in the problem statement and calculations, as well as the need for clarity on the integration setup and the treatment of the electric field's direction.

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Homework Statement


If V=2x^2+6y^2 V in free space, find the energy stored in a volume defined by -1<x<1,-1<y<y, and -1<z<1. (BTW the < are suppose to less than or equal to or greater than or equal to)


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The Attempt at a Solution


I am not really sure of the formula here but I assume it needed at triple integral as it gave me 3 sets of limits and a function. Furthermore it asked for a volume so it was the only thing I could think of I relize it is probably wrong but am I even close? If not what formula should I use?
 

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A triple integral is good, but you have to find out what to integrate first.

I'm not sure how to interpret "V=2x^2+6y^2 V", but it looks like an electric potential, so we have an electric field in that volume. What is the energy density of an electric field?
Are you sure there are no units involved?
 
Oh sorry yah the V stands or volts so yes it is electric potential. (1/2)(epsilon)E^2 equals energy density of an electric field
 
Good, now you can use that to solve the problem.
 
So what your saying is that E would equal electric potential here or should I use dV=-Es(dS)
 
Thus I would take the derivative of the voltage
 
DODGEVIPER13 said:
Thus I would take the derivative of the voltage
That's the idea, right.
 
Well what should I derive the function by x or y I'm guessing x
 
DODGEVIPER13 said:
Well what should I derive the function by x or y I'm guessing x

x,y, and z, actually. You want to determine the electric field, a vector quantity, by finding the gradient of the potential. Presumably you'll then use the magnitude of this vector in your energy density evaluation (triple integral over the volume).
 
  • #10
Ah ok man thanks so partial derivative with respect to x y and z then integrate that three times with the limits given got it will do later on
 
  • #11
Well the gradient is (4x,12y) if I integrate it straight I get 0 which can't be right . I don't know if it is mathematically legal for me to e the agnitude with the x and y arable still in it but ill try anyway see if that does the trick I gets 101.19288
 
  • #12
Remember that you have to square its magnitude:
DODGEVIPER13 said:
Oh sorry yah the V stands or volts so yes it is electric potential. (1/2)(epsilon)E^2 equals energy density of an electric field
 
  • #13
6e-9
 
  • #14
I got the magnitude to be sqrt(4^2+12^2) which is sqrt(160)=E so then E^2=160 which when I integrate I get 6e-9
 
  • #15
Where epsilon is 8.85e-12
 
  • #16
(4x,12y) depends on x and y, and so does the magnitude. In addition, you are missing the z-component here (it is zero, but you have to include it to get a vector with 3 components).
 
  • #17
I get E=sqrt((4x)^2+(12y)^2)=sqrt(16x^2+144y^2) then E^2=16x^2+144y^2 after integrating I get (1/2)epsilon(1280/3)=2e-9
 
  • #18
Am I ok not trying to rush you just wondering?
 
  • #19
You can check calculations with a computer, so what are you waiting for?
The answer has missing units, but the problem statement has the same problem.
 
  • #20
Ok thanks man yah u are right wolfram said it was ok. I guess I said it wrong did I have the integral set up correctly so that when I performed the integration it would be correct?
 
  • #21
That looks good.
 
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  • #22
Ok thanks man
 
  • #23
oh hey I know this post is old but I caught a mistake I believe since dV=-E dot dS then the answer shourld be negative 2E-9 also what should the units be Joules?
 
  • #24
You square the electric field, its direction does not matter, and the energy density is always positive.
 
  • #25
Ah yah your right didnt consider that it was squared.
 

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