What is the magnitude and direction of the electric field

  • #1

Homework Statement


What is the magnitude and direction of the electric field at the position indicated by the dot in the figure below? Give your answer in component form in the blanks below.
phys2pic.jpg


What is the x-component of the electric field at the indicated point?

Homework Equations


kq/r^2

The Attempt at a Solution


Electric field from 1 = [(8.99x10^9)(-10x10^-9)]/((0.03)^2) = -99888.8 Nc
Electric field from 2 = [(8.99x10^9)(10x10^-9)]/((0.05830951895)^2) =26441.17 Nc
Electric field from 3 = [(8.99x10^9)(-10x-5^-9)]/(0.05^2) = -17980 Nc


Total magnitude of the net electric field = sqrt((-99888.8)^2+(26441.17)^2+(-17980)^2) = 104881.8685 Nc

Now I am confused as to how to find my angle for the direction at the point in the picture. there are three different particles having an effect, if there where one id know how to find the correct angle to find the x component, but I am unsure how to get my angle here
 

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Answers and Replies

  • #2
I added up the vectors graphically and found the magnitude and got an angle of 252 degrees in quadrant 3. does this look correct?
 
  • #3
Doc Al
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Electric field from 1 = [(8.99x10^9)(-10x10^-9)]/((0.03)^2) = -99888.8 Nc
Electric field from 2 = [(8.99x10^9)(10x10^-9)]/((0.05830951895)^2) =26441.17 Nc
Electric field from 3 = [(8.99x10^9)(-10x-5^-9)]/(0.05^2) = -17980 Nc
Each of these field contributions has a direction. See if you can figure out each one's x and y component.
 
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  • #4
Each of these field contributions has a direction. See if you can figure out each one's x and y component.
hey please see above, I added the vectors graphically on graphing paper and got 252 degrees
Each of these field contributions has a direction. See if you can figure out each one's x and y component.
So im imagining a unit circle around the point, charge 1 has an angle of 90 degrees, charge 2 has an angle of arctan(3/5) = 30 degrees and charge 3 has an angle of 0 degrees, but then how do I find the angle on the particle?
 
  • #5
Doc Al
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Try this: The field from the -10 nC charge (at the point in question) acts in what direction? What would be its x and y components?
 
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  • #6
Try this: The field from the -10 nC charge (at the point in question) acts in what direction? What would be its x and y components?
The x component would be cos(270) = 0 and the y component would be sin(270) = -1
 
  • #7
Doc Al
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The x component would be cos(270) = 0 and the y component would be sin(270) = -1
Almost: Tell me in words which direction that field points.
 
  • #8
Almost: Tell me in words which direction that field points.
the direction of the field from charge 1 onto point P points straight down
 
  • #9
Doc Al
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the direction of the field from charge 1 onto point P points straight down
Why down? (The charge is negative.)
 
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  • #10
Why down? (The charge is negative.)
am I supposed to assume that the test charge is positive?? then up. wasn't sure about what sign the test charge was
 
  • #11
Doc Al
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am I supposed to assume that the test charge is positive?? then up. wasn't sure about what sign the test charge was
No test charge is needed or mentioned. You're finding the field at a point.
 
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  • #12
No test charge is needed or mentioned. You're finding the field at a point.
oh god im confused now
 
  • #13
Doc Al
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Hint: The field from a positive charge points in which direction: Towards or away from the charge?
 
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  • #14
I thought you need a test charge to find the field at a point
 
  • #15
Hint: The field from a positive charge points in which direction: Towards or away from the charge?
away
 
  • #16
Doc Al
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  • #17
Nope. (But it can be useful to imagine one there.)


Good. And for a negative charge?
towards, so it would be pointed upwards
 
  • #18
Doc Al
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towards, so it would be pointed upwards
Exactly. So the y-component would be what?

Now do the same thinking for the other charges.
 
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  • #19
Exactly. So the y-component would be what?

Now do the same thinking for the other charges.
so the y component would be 99888.8sin(90) = 99888.8 Nc
 
  • #20
oh so I just sum the components
 
  • #21
So do you agree the net component for x is -4693 nC
 
  • #23
I got 1.62/2 points isn't the unit nanocolumn?
 
  • #24
Doc Al
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I got 1.62/2 points isn't the unit nanocolumn?
I haven't checked your numbers. The proper unit for electric field is N/C (Newtons per Coulomb).
 
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