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Electric field at the center of the equilateral triangle

  • Thread starter mlostrac
  • Start date
  • #1
83
0

Homework Statement


Consider an equilateral triangle of side 15.6 cm. A
charge of +2.0 µC is placed at one vertex and
charges of –4.0 µC each are placed at the other
two, as shown in the diagram to the right.
Determine the electric field at the centre of the
triangle.




I've found the distance from each point to the centre. Now do I just find the electric field of each compared to the centre and add all three together?
 

Answers and Replies

  • #2
461
8
That is the way.
 
  • #3
83
0
Ok, I tried it and here's what I got:

Distance from the centre to the top vertice of the triangle (with a charge of +2.0 uC) = 8.4 cm
Distance from each of the two bottom vertices to the centre (each with a charge of -4.0 uC) = 3.2 cm

E = kQ1/r^2 + kQ2/r^2 + kQ3/r^2
= k (Q1/r^2 + Q2/r^2 + Q3/r^2)
= 9 x 10^9 ( 0.002/0.084 + 0.004/0.032 + 0.004/0.032)
= 2.46 x 10^9 N/C

Does that look like I did everything right? And, does the direction matter, or is that implied based on the negative and positive charges...
 
Last edited:
  • #4
195
1
Electric field is a vector.
You cannot add up field like you add up numbers.
Draw the field vector at the center due to each of the charges. Then you'll have to resolve the fields into x & y components. You can then add up the x components & the y components.
 

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