Resultant electric field magnitude and direction

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Homework Help Overview

The problem involves calculating the resultant electric field at the center of a square with four point charges located at its corners. The charges are specified in nanocoulombs, and the medium is air. Additionally, the force on a smaller charge placed at the center is also of interest.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of superposition to calculate the electric fields from each charge and combine them as vectors. There are attempts to clarify the calculations for the x and y components of the electric field.

Discussion Status

Some participants provide feedback on the calculations, noting a potential oversight in the y-component summation. There is an acknowledgment of the original poster's method, with suggestions for refinement rather than a complete resolution.

Contextual Notes

The original poster expresses a lack of prior physics experience, which may influence their understanding of the concepts involved. There is a focus on ensuring the calculations are correctly executed without providing direct solutions.

ledwardz
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Homework Statement



a square of sides 10cm has four charges at each corner starting clockwise from top right they are 5,10,6 and 3nC. Determine the resultant electric field including magnitude and direction at the centre of the square. Owww and the medium is air. Finally what is the force on a 6pC placed at this point.

i should probably point out I've never done physics in my life be4 u give me grief... take pity


Homework Equations



okay first part I am using

kQ/R^2
k= 1/4pi* E0*Er
E0= 8.854*10^-12
Er= 1.005
my value of r is 7.07cm which = 0.0707m

sooooo i have 4 electric fields of;

8945.723 Cm^-1
17892.28 ...
10735.37 ...
5367.68 ...

solving as vectors...

(x)= 5367.68cos45 + 10735.37cos45 - 17892.28cos45 - 8945.723cos45
= 7590.55 to the left

(y)= 10735.37sin45+17892.28sin45 - 8945.723sin45 - 5367.68sin45
= 2530 upwards

gives me a magnitude of 8000 Cm^-1
at an angle of 18.43

then the force is

q*E = 6*10^-9 *8000 = 4.8*10^-5N

Thanks for any help, Lee.:cry:
 
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use superposition of electric fields, adding them as vectors. you will find the net field's magnitude as well as direction easily.
 
supratim1 said:
use superposition of electric fields, adding them as vectors. you will find the net field's magnitude as well as direction easily.



I think that is what i did, have i got the right idea?
 
ledwardz said:
...

solving as vectors...

(x)= 5367.68cos45 + 10735.37cos45 - 17892.28cos45 - 8945.723cos45
= 7590.55 to the left  I agree.

(y)= 10735.37sin45+17892.28sin45 - 8945.723sin45 - 5367.68sin45
= 2530 upwards  I think you missed adding the 10735.37sin45°. Otherwise it looks good.

Re-Do what follows.

gives me a magnitude of 8000 Cm^-1
at an angle of 18.43

then the force is

q*E = 6*10^-9 *8000 = 4.8*10^-5N

Thanks for any help, Lee.:cry:

supratim1 said:
use superposition of electric fields, adding them as vectors. you will find the net field's magnitude as well as direction easily.
Hello Lee.

See my comments in red above.

Your method looks good. You appear to have merely missed one term in the y component!
 
SammyS said:
Hello Lee.

See my comments in red above.

Your method looks good. You appear to have merely missed one term in the y component!



cheers man, appreciate the help. good to know i am doing summit right for once:approve:e.
 

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