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Resultant electric field magnitude and direction

  • Thread starter ledwardz
  • Start date
  • #1
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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 ive never done physics in my life be4 u give me grief....... take pity


Homework Equations



okay first part im 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:
 

Answers and Replies

  • #2
supratim1
Gold Member
279
1
use superposition of electric fields, adding them as vectors. you will find the net field's magnitude as well as direction easily.
 
  • #3
7
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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.


I think that is what i did, have i got the right idea?
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,312
1,001
...

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:
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!
 
  • #5
7
0
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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