What is the magnitude and direction of the electric field

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SUMMARY

The magnitude and direction of the electric field at a specified point due to three charges were calculated using Coulomb's law, represented by the equation kq/r². The contributions from each charge were computed, resulting in a total electric field magnitude of approximately 104,881.87 N/C. The direction was determined graphically, with an angle of 252 degrees in the third quadrant. The discussion clarified the importance of understanding the direction of electric fields from both positive and negative charges, emphasizing that the electric field from a positive charge points away while that from a negative charge points towards the charge.

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  • Coulomb's law for electric fields (kq/r²)
  • Vector addition of forces
  • Understanding of electric field directionality
  • Basic trigonometry for angle calculations
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  • Learn about electric field lines and their representation
  • Explore the concept of superposition in electric fields
  • Investigate the effects of multiple charges on electric field strength and direction
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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 NcTotal 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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I added up the vectors graphically and found the magnitude and got an angle of 252 degrees in quadrant 3. does this look correct?
 
isukatphysics69 said:
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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Doc Al said:
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
Doc Al said:
Each of these field contributions has a direction. See if you can figure out each one's x and y component.
So I am 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?
 
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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Doc Al said:
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
 
isukatphysics69 said:
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.
 
Doc Al said:
Almost: Tell me in words which direction that field points.
the direction of the field from charge 1 onto point P points straight down
 
isukatphysics69 said:
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
Doc Al said:
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
isukatphysics69 said:
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
Doc Al said:
No test charge is needed or mentioned. You're finding the field at a point.
oh god I am confused now
 
  • #13
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
Doc Al said:
Hint: The field from a positive charge points in which direction: Towards or away from the charge?
away
 
  • #16
isukatphysics69 said:
I thought you need a test charge to find the field at a point
Nope. (But it can be useful to imagine one there.)

isukatphysics69 said:
away
Good. And for a negative charge?
 
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  • #17
Doc Al said:
Nope. (But it can be useful to imagine one there.)Good. And for a negative charge?
towards, so it would be pointed upwards
 
  • #18
isukatphysics69 said:
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
Doc Al said:
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
 
  • #22
YES I GOT IT!
 
  • #23
I got 1.62/2 points isn't the unit nanocolumn?
 
  • #24
isukatphysics69 said:
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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