# Electric Field Question

1. May 24, 2010

### SD2014

1. The problem statement, all variables and given/known data
A point charge q = 2.7 x 10^-6 C is placed at each corner of an equilateral triangle with sides of .11 m in length. What is the magnitude of the electric field at the midpoint of any of the 3 sides of the triangle.

2. Relevant equations
F = kqq/r^2
E = F/q

3. The attempt at a solution
F = k(2.7x10^-6)^2/.055^2
F = 21.66 N

Since there are two forces acting at this point
2 x 21.66 = 43.22 N is the total force.

E = F/q
43.22 N/2.7x10^-6C
= 16007407 N/C

2. May 24, 2010

### jyothsna pb

the intensity due to the 2 charges on the same side of the triangle ( where d point is taken) cancel out since the charges are of same nature ( forces act in opposite directions). so the net field is contributed only by the charge on opposite vertex.

3. May 25, 2010

### thebigstar25

I always believe that when you have a problem such this, it is better to draw a simple figure to be sure that what you are doing is right ..

so for your problem see the following figure:

http://img28.imageshack.us/img28/1414/64786042.jpg [Broken]

figure (1) represents your problem, you have identical charges at the croners of the triangle..

lets consider the situation (figure (2) ) where we want to find the field at the midside (the one on the left) .. notice carefully the directions of the fields from each charge , what can you conclude?

Last edited by a moderator: May 4, 2017