Energy in a volume

  • #1
DODGEVIPER13
672
0

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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Answers and Replies

  • #2
36,020
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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?
 
  • #3
DODGEVIPER13
672
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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
 
  • #5
DODGEVIPER13
672
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So what your saying is that E would equal electric potential here or should I use dV=-Es(dS)
 
  • #6
DODGEVIPER13
672
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Thus I would take the derivative of the voltage
 
  • #8
DODGEVIPER13
672
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Well what should I derive the function by x or y I'm guessing x
 
  • #9
gneill
Mentor
20,945
2,886
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
DODGEVIPER13
672
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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
DODGEVIPER13
672
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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
36,020
12,919
Remember that you have to square its magnitude:
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
DODGEVIPER13
672
0
6e-9
 
  • #14
DODGEVIPER13
672
0
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
DODGEVIPER13
672
0
Where epsilon is 8.85e-12
 
  • #16
36,020
12,919
(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
DODGEVIPER13
672
0
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
DODGEVIPER13
672
0
Am I ok not trying to rush you just wondering?
 
  • #19
36,020
12,919
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
DODGEVIPER13
672
0
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?
 
  • #22
DODGEVIPER13
672
0
Ok thanks man
 
  • #23
DODGEVIPER13
672
0
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
36,020
12,919
You square the electric field, its direction does not matter, and the energy density is always positive.
 
  • #25
DODGEVIPER13
672
0
Ah yah your right didnt consider that it was squared.
 

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