Electric Field at Center of Equilateral Triangle

[Myr73]

Consider an equilateral triangle of side 15.6 cm. A charge of +2.0uc is placed at one vertex and charges of -4.0C uc 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

ANgle= 60 sides--> d1= d2=d3=0.156m q1=2X10^-6C q2=q3=-4X10^-6C
E(center)=?

E=kQ/r^2 {E= E1+E2+E3

Let q1 be the charge on top, q2 bottom left, q3 bottom right. Side d1, left side, side d2 , right side and side d3, bottom side of triangle.

What I am having the most trouble with here, is finding the center of the triangle, in order to determine r. I think that the center would be found in a picture, by drawing from the midpoint of one side to the opposite corner, doing this for every side , and then the point where all three lines meet this would be the center. However I am having trouble knowing how to calculate this.

Can you please help?

 

[Nathanael]

I think that the center would be found in a picture, by drawing from the midpoint of one side to the opposite corner, doing this for every side , and then the point where all three lines meet this would be the center.

Use some trigonometry.

I'll call the (perpendicular) height from one of the bases to the center "h"

I'll call the side length "s" (it should be the same for all 3 sides, since it's "equi-lateral")

If you picture it, you should see a relationship between s, h, and tan(θ)
(where θ is half of one of the triangle's angles, since it was bisected by the lines you mentioned)

 

[HallsofIvy]

If you set the triangle in a coordinate system, so that the vertices are $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, it is not difficult to show that the center has coordinates equal to the average of those: $(x_1+ x_2+ x_3)/3$, $(y_1+ y_2+ y_3)/3$.

 

[Chestermiller]

The bisectors of the three angles meet at the center. Does that help?

Chet

 

[Myr73]

Yes-Thanks :), all of it does!

Oh but wait, for the explanation with h- How would I know what h would be?

 

[Nathanael]

Oh but wait, for the explanation with h- How would I know what h would be?

You mean mine? Sorry I wasn't very clear

h would be $\frac{s}{2}tan(30°)$ but that's the distance from the center to the side, not the center to the vertex

(The center-to-vertex length can be found similarly)

 

[Myr73]

aghh- yap makes perfect sense now, thank you

 

[Myr73]

Ok, so one quick question- What ways do I make the charges Q go, or the energy I suppose based on the charges. So Q1 is positive , Q2 and Q3 are negative. I'm wondering if I need to take into consideration the force they have on each other, or since its based on the center, maybe not. I was thinking that the charges Q2 and Q3 (negatives) would go towards the center, and Q1 would go towards the outside. Is this correct? Or would it be viseversa?

 

[Myr73]

or since "if q are positive, F and E ar in the same direction. If q is negative, F and E are in the opposite directions" Would the forces for the negative charges point outside and therefore there energy point inside, and the forces of the positive force point inside, and therefore it's energy point inside.??

 

[Infinitum]

Ok, so one quick question- What ways do I make the charges Q go, or the energy I suppose based on the charges. So Q1 is positive , Q2 and Q3 are negative. I'm wondering if I need to take into consideration the force they have on each other, or since its based on the center, maybe not. I was thinking that the charges Q2 and Q3 (negatives) would go towards the center, and Q1 would go towards the outside. Is this correct? Or would it be viseversa?

For your original question above there is no need to consider the motion of the charges, as you are dealing with electrostatics and electric field at the moment. But yes, the forces acting on Q1, Q2, Q3 due to each other can be found quite easily using Coulomb's law, which can indicate the direction of acceleration of the charges. The resultant direction of the acceleration of any charge would be the direction of the force on it due to the other charges, which, as you said, means the negative charges would tend to move away 'outside' the triangle boundary, where as the positive charge would tend to be attracted towards the center of the triangle.

or since "if q are positive, F and E ar in the same direction. If q is negative, F and E are in the opposite directions" Would the forces for the negative charges point outside and therefore there energy point inside, and the forces of the positive force point inside, and therefore it's energy point inside.??

I'm not entirely sure what you are trying to ask here. Energy is a scalar quantity, it doesn't 'point' anywhere. The electric field does have a direction however, and it is defined as the direction a positive charge would be pushed in while placed in that field. Based on that what can you conclude about the direction of the field in the above case?

 

[Myr73]

Sorry, I should have said, electric field, witch direction is the electric field E pointing. I understand that I have to break that up into x and y, but I have to know witch directions they would be pointing.
And,actually for part 1, I had said that I thought the E from the negative charge would point towards the inside not the outside. So I am still confused , lol

 

[Infinitum]

What you need is the direction of the electric field at the center of the triangle. There doesn't arise a question of inside or outside here. Draw a free body diagram to help understand better. Now, iterate through each of the charges, for example, Q1(the positive charge) first. Which direction would this Q1 effect unit positive charge(for simplicity) placed at the center of the triangle? Similarly think how the negative charge Q2, and then Q3, on both vertices would effect the unit positive charge placed at the center, and this should give you the resultant direction of the field.

 

[Chestermiller]

Ok, so one quick question- What ways do I make the charges Q go, or the energy I suppose based on the charges. So Q1 is positive , Q2 and Q3 are negative. I'm wondering if I need to take into consideration the force they have on each other, or since its based on the center, maybe not. I was thinking that the charges Q2 and Q3 (negatives) would go towards the center, and Q1 would go towards the outside. Is this correct? Or would it be viseversa?

In an electrostatics problem like this, it is assumed that the three charges Q1, Q2, and Q3 are held in place by some unspecified forces that prevent them from moving. You are then asked to determine the electric field intensity vector at some other location in space (in your problem, the center of the triangle) by imagining that you place a small positive test charge at that location (that has negligible effect on the electric field) and asking what resultant force the three charges would exert on the test charge. This is another way of saying what Infinitum is describing. Did they not cover this in your textbook or course?

Chet

 

[Myr73]

umm, I don't remember this being covered-

 

[Chestermiller]

umm, I don't remember this being covered-

Please go back to your textbook and see if you can find it there. It's pretty basic to understanding this problem.

Chet

 

[Myr73]

it kinda makes more sense now, I am still not sure though about the statement ""if q are positive, F and E are in the same direction. If q is negative, F and E are in the opposite directions" " Which is out of the given physics manual-
and then also if there were a positive charge in the middle, this would affect the direction of the forces no?

 

[Myr73]

ok, ill check it out

 

[Myr73]

ok umm does this mean that E2x and E2y would be negative,(pointing out of the triangle) E3x would be positive,E3y would be negative,(also pointing out,towards charge 3) and E1x is N)./A and E1y is negative(poiting towards the middle )?