Coulomb Forces

  • Thread starter azizlwl
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  • #1
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Homework Statement


Find the force on a point of charge of 30μC at (0,0,5)m due to a 4m square int he z=0 plane between x=±2m and y=±2m with a total of 500μC, distributed uniformly.

Homework Equations




The Attempt at a Solution


R=-x,-y,+5z
dQ=ρdydx=500/4 dydx

dE=dQ(5az) / (4πε0((x2+y2+25)3/2 )
x and y axis cancel

How to do integral with x2 and y2 in the denominator?

Ans. 4.66azN
 
Last edited:

Answers and Replies

  • #2
Simon Bridge
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... use the relationship of dQ to dx and dy to make a double integral.
 
  • #3
1,065
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... use the relationship of dQ to dx and dy to make a double integral.
... use the relationship of dQ to dx and dy to make a double integral.
 
  • #4
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Actually I really don't understand the question. Area enclosed by x=±2 and y=±2 will be 16m square. What does it means by" due to a 4m square "?
 
  • #5
gneill
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Actually I really don't understand the question. Area enclosed by x=±2 and y=±2 will be 16m square. What does it means by" due to a 4m square "?
Draw a square 4 m per side. Distribute 500 μC uniformly on the edges (so each side of the square is a line segment with 500/4 μC on it.
 
  • #6
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F1d=3x5x9/16 ∫-22 dy (√29d -yay)/(29+y2)3/2
F1d=8.43 x 0.129 =1.087
The integral value using app integral calculator.
F1z=1.087 x 5 /√29 = 1.009.

Total force= 4Naz less than the answer..
 
  • #7
gneill
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F1d=3x5x9/16 ∫-22 dy (√29d -yay)/(29+y2)3/2
F1d=8.43 x 0.129 =1.087
The integral value using app integral calculator.
F1z=1.087 x 5 /√29 = 1.009.

Total force= 4Naz less than the answer..
I don't understand what your last line is trying to convey. Can you explain in more detail?
 
  • #8
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4 segments, so I added all that z-axis component. X and y axis cancelled. The answer is 4.66azN
 
  • #9
gneill
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So your result is about 4 N and the answer given in the text is 4.66 N ?

I'm inclined to agree with your result over that of the book.
 

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