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Electric Field at the Center of a Triangle

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Homework Statement



Given an equilateral triangle with sides 20-cm, with point charges at each point:
q1 at the top with a charge of +4.0uM,
q2 at the bottom right with a charge of -4.0uM, and
q3 at the bottom left with a charge of +4.0uM.

k is assumed to be exactly 9.00x109 N-m2/c2

Question: What is the electric field at the center of the triangle?


Homework Equations



E = (kq)/r2

The Attempt at a Solution



First and foremost, I employed a little geometry find the center of the triangle:

r = sqrt( (0.20-m)2 - (0.10-m)2 ) / 2
r = 0.866-m, rounded to
r = 0.90-m

E1 = (9.00x106 N-m2/c2)(4.0x10-6C) / (0.09-m)2
E1 = 4.4x106 N/C

The magnitudes of all three charges are equivalent, so:

E1 = E2 = E3

So the x and y components would be:

Ex = (4.4x106 N/C)cos30° + (4.4x106 N/C)cos330°
Ex = 7.6 x 106 N/C

Ey = (4.4x106 N/C)sin270° + (4.4x106 N/C)sin30° + (4.4x106 N/C)sin330°
Ey = -4.4x106 N/C

E = sqrt[ (7.6 x 106 N/C)2 - (4.4x106 N/C)2 ]
E = 6.2 x 106 N/C

So, my answer is 6.2 x 106 N, but the book answer key shows 5.4 x 106 N/C. That is quite a significant difference, but I can't seem to find where I'm making a mistake. It's driving me mad!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
ehild
Homework Helper
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1,876


r = sqrt( (0.20-m)2 - (0.10-m)2 ) / 2
r = 0.866-m, rounded to
r = 0.90-m



Check your geometry. How far are the corners of an equilateral triangle from the centre?

You rounded r too much, just at the beginning of your calculations.
The value of r has got 4% error because of rounding. It is squared, that doubles the error to 8% which increases further during the rest of the calculations.

You must not round so much during the calculations. Keep 3-4 significant digits.

You have r2 in the formula for the electric field intensity. So you do not need the square root at all and.


ehild
 
  • #3
4
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I must be daft!

And I made a typo, it's 0.09 and not 0.90. But that's beyond the point!

The distance from one corner of an equilateral triangle is equal to a hypotenuse made from a right triangle with one leg half the length of teh sides of the triangle and the other leg equal to 1/3 the value of the distance from one corner to the midpoint of the opposing side!

SO

r2 = [ sqrt( (0.20m)2 - (0.10m)2 ) / 3 ]2 + (0.1m)2
r2 = 0.0133 m

Thank you very much, ehild! :-D
 

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