# Homework Help: Center E field of an equilateral triangle.

1. Oct 18, 2014

### RyanBruceX

y/SUB]1. The problem statement, all variables and given/known data
Determine the electric field at the center of the triangle. All sides are 15.6 cm. Vetex is 2.0 uC (Q1) while the bottom points are both -4.0 uC (Q2 and Q3).

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

3. The attempt at a solution

Ecenter = E1 + (2)E2 = ?

First calculate midpoint of triangle by splitting it down the middle, resulting in the top left angle being 30 degrees. To determine the height of the triangle I used cos30L where L = 0.156 m = 0.135 m. So midpoint will be half this = 0.0657 m.

Now my first question is the E from Q1 going to simply be E = k2.0 x 10^-6/ 0.135^2? Or because Q1 (top of triangle) acts at angles towards Q2 and Q3 the E field felt from Q1 on p (middle) will be 2Ey after cancelling Ex due to symmetry?

If going with my first solution we now have Ec = 9.876 x 10^5 N/C + 2Ey = ?

Distance from Q2/Q3 to center = square of (0.078^2 + 0.0675^2) = 0.103 m.
E = k(-4.0 x 10^-6)/0.103^2 = -3.39 x 10^-6. So Ey = sin30E = -1.696 x 10^6.

Net E middle of triangle = E1 + 2Ey = 9.876 x 10^5 + (2)1.696 x 10^6 = 4.379 x 10^6 N/C?
Unless the top charge has to be broken into components I think this is right? thanks.

2. Oct 18, 2014

### ehild

No. The contributions of Q2 and Q3 to the electric field are not the same, the directions are different. The electric field is a vector. Apply vector addition to get the resultant from the contributions of the individual charges.

The centre of the triangle is the point which is at equal distance from all vertexes. It is not at the midpoint of the height.

ehild

3. Oct 19, 2014

### RyanBruceX

Okay, I do not know ho to find the middle point of the triangle. But once I figure that out am I correct in thinking the net E will be
E = k(+2uC)/distance top vertex to center^2 + E = (2)(sin30)(k-4uC/distance from bottom apex to center^2 = ? I see that both E fields move -y so their addition will make E greater in the -y direction than either one alone.

Or do I have to double the y component of the +2uC charge as well? Thanks

4. Oct 19, 2014

### Staff: Mentor

You dropped a perpendicular from the top vertex down to the triangle's base, thus bisecting the angle at the top vertex. Do the same for the other two vertices, dropping perpendiculars to their opposite sides. Can you spot the symmetry that results? Where might you find the center of the triangle?

5. Oct 19, 2014

### RyanBruceX

Ok, so when I do that I can visualize the center, but not know its physical location. So now that I see center it looks as though from the left bottom vertex I can calculate the height from base to midpoint using tan30 (0.5 x 0.156m) = 0.045 m? And Q2 and Q3 are 0.09 m from vertex to midpoint. Q1 to midpoint = 0.135 - 0.045 = = 0.09 m which makes sense due to symmetry.

SO now the solution. E = k(+2uC)/0.09^2 + (2)sin30(k-4uC/0.09^2) = 2.22 x 10^6 + 4.44 x 10^6 = 6.7 x 10^6 N/C (-y direction).. Does this look right?

6. Oct 19, 2014

### ehild

That is right.

E = k(+2uC)/0.09^2 + (2)sin30(k-4uC/0.09^2) is wrong. In what direction does the field of the 2uC charge point?
(k-4uC/0.09^2) means that you subtract (k-4uC/0.09^2) from k which has no sense.
Accidentally, the end result is correct.

ehild

7. Oct 19, 2014

### RyanBruceX

I should have put -0.09^2 for the distance for the +2.0uC equation. I intuitively knew it was going to be a the sum of the fore magnitudes and in the negative direction. Thank you very much for your help!! Greatly appreciated!