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Gauss's Law - Cube with point charge at centre

  • Thread starter roman15
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  • #1
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Homework Statement


a)A charge Q=10nC is placed at the origin, which is the centre of a cube with side lengths a=1cm whose faces are perpendicular to the x,y and z axis. What is the total flux through the box and what is the average of the perpendicular component of the electric field on the right hand face centred on (0, a/2, 0)?

b)Now a second charge, -Q, is placed at (0,a,0). Now what is the total flux and the average fo the perpendicular component of the electric field at (0,a/2,0)?


Homework Equations


flux=Q/eo
so i was able to calculate the flux easily, but i dont understand what is meant by the average perpendicular component of the electric field
the only thing i can think of it that EA=Q/eo, so i just solve for E? is that right

for the second part, wouldnt the total flux still be the same, because the second charge isnt enclosed in the cube


The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
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hi roman15! :smile:

(have an epsilon: ε :wink:)
flux=Q/eo
so i was able to calculate the flux easily, but i dont understand what is meant by the average perpendicular component of the electric field
the only thing i can think of it that EA=Q/eo, so i just solve for E? is that right
i think they just mean the flux though that face, divided by the area :wink:
for the second part, wouldnt the total flux still be the same, because the second charge isnt enclosed in the cube
yup! :biggrin:
 
  • #3
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oh so then it would be 1/6 of the flux divided by the area of that face!
oh and then wouldnt the average perpendicular component for part b also be the same? because the flux and area are the same
 
  • #4
tiny-tim
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yes! :smile:

and yes, the extra would be the same :wink:
 
  • #5
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oh and then wouldnt the average perpendicular component for part b also be the same? because the flux and area are the same
No. In part a equal flux passes through each face of the cube, but not in part b! So you can't just take the total flux through the cube and divide by 6 to get the flux through the specified face. You should figure out the flux through that face due to the first charge, and the flux due to the second charge, and sum them up to get the total flux through that face (then divide by the area of the face to get the average perpendicular electric field across the face [you do this because the electric flux per unit area through a surface is defined as the component of the electric field perpendicular to the surface]).
 
  • #6
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oh ok
but the second charge is equal but opposite to the first charge, so wouldnt the flux from the second charge cancel out with the flux from the first charge, so then the perpendicular electric field would be zero
 
  • #7
tiny-tim
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yes … if i'm reading the question right, the charges are equal and opposite, and the displacements are equal and opposite (ie it's a mirror! :wink:), so the fluxes are the same :smile:
 
  • #8
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but then that doesnt make sense that the field between two opposite charges would be zero
 
  • #9
tiny-tim
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oops!

but then that doesnt make sense that the field between two opposite charges would be zero
oops! I got confused :confused:

i don't know why i used "displacement", it's irrelevant :redface:

i should have said: "the charges are equal and opposite, and the distances are equal (ie it's a mirror! :wink:), so the fluxes are the equal and opposite :smile:"

sorry! :redface:
 
  • #10
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its alright lol, but im still confused, because if the total flux is zero then that means there is no electric field between the charges, but there has to be...
 
  • #11
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The total flux through the face isn't zero. Draw the setup and you will see that the fluxes from the two charges are in the same direction through the face, and so add together (instead of cancelling each other).
 
  • #12
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oh because one is positive and one is negative, the field lines are pointing in the same direction
so would the total flux=2Q/eo?
and then the electric field would be that divided by the area?
 
  • #13
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Q/e0 tells you the flux through a closed surface in terms of the total charge enclosed by that surface. But here we want the flux through a square, which isn't a closed surface.

However, you can make the argument tiny-tim is talking about: since the charges have equal magnitudes, and are placed symmetrically at equal distances from the face, the flux through the face from each charge is equal. And you already calculated the flux through that face from one of the charges.
 
  • #14
tiny-tim
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roman15 said:
oh because one is positive and one is negative, the field lines are pointing in the same direction
so would the total flux=2Q/eo?
and then the electric field would be that divided by the area?
yes the fluxes are opposite, but they're going through the surface in opposite directions, so they add :smile:
 
  • #15
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ok thank you guys so much!! I really appreciate it
 

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