What is the magnitude and direction of the electric field

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Homework Help Overview

The discussion revolves around determining the magnitude and direction of the electric field at a specific point influenced by multiple charges. The problem involves calculating the electric field components using the formula kq/r² and understanding the vector nature of electric fields.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss calculating the electric field contributions from three different charges and express confusion about determining the resultant angle and direction of the electric field. There are attempts to break down the contributions into x and y components, and questions arise regarding the necessity of a test charge in the analysis.

Discussion Status

Participants are actively engaging with the problem, exploring different methods to calculate the electric field components and questioning the assumptions about charge signs and directions. Some guidance has been provided regarding the direction of electric fields from positive and negative charges, but no consensus has been reached on the final calculations.

Contextual Notes

There is uncertainty regarding the sign of the test charge and its relevance to the problem, as well as the interpretation of angles related to the electric field vectors. Participants are also clarifying the units of measurement for electric fields.

isukatphysics69
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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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