# Charges on an equilateral triangle

1. Apr 6, 2012

### 4real4sure

1. The problem statement, all variables and given/known data
Charges Q, Q, and q lie on the corners of an equilateral triangle with sides of length a. Charge q lies on the top corner with Q and Q on the left and right corners.

(a) What is the force on the charge q?

(b) What must q be for E to be zero half-way up the altitudeat P?

2. Relevant equations

F=(1/4πε) * (q1q2/r2)

E = (1/4πε) * (q/r2)

3. The attempt at a solution

(a) F = qQ / (4πεa2) iy

(b) Eq = q / (4πε (3/16 a2)) iy

and EQ = Q / (4πε (7/16 a2)) iy

I just wanted to know if the attempts are in proportion to the required?

2. Apr 6, 2012

### collinsmark

No, that's not quite right.

You are correct that the net force is completely in the $\hat \imath_y$ direction. That is because the $\hat \imath_x$ components of each force from the Q charges cancel.

But the $\hat \imath_y$ force components do not lead to your answer.
Those electric field calculations are not quite right either.

3. Apr 6, 2012

### 4real4sure

I tried to do it several times based on may understanding but for some reason I am so not able to reach to the solution. Any hint?

4. Apr 6, 2012

### Staff: Mentor

Did you begin with a diagram of the triangle, labeling all sides in cluding the altitude?

If you calculate the force that one Q exerts upon q along a side a, then what is the y-component of that force? Hint: you can determine the trigonometric functions (sin, cosine) of the angle using an appropriate ratio of sides.

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5. Apr 6, 2012

### collinsmark

For each charge Q at the bottom, you can find the magnitude of the force from that particular charge on charge q using

$$\vec F = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{r_2} \hat \imath_r.$$

where $\hat \imath_r$ points in the direction from the particular Q to q.

But the $\hat \imath_r$ for one charge is different from the $\hat \imath_r$ of the other charge, because they point in different directions. So you can't just add them simply. For each case, you need to break up $\hat \imath_r$ into its $\hat \imath_x$ and $\hat \imath_y$ components.

That might involve something like

$$F_x = F\cos \theta$$
$$F_y = F \sin \theta$$
(This relationship is just for example purposes, the particular relationship may depend on the particular problem/particular pair of charges).

Or if you don't want to use trigonometric functions, you can do it using the Pthagorean theorem (which works with this problem). If a is the side of an equilateral triangle, and you vertically bisect the triangle down the middle, forming two isosceles triangles, then the base of each isosceles triangles is a/2. What does Pthagorean theorem tell you about the height of the triangle relative to a?

6. Apr 7, 2012

### 4real4sure

I tried and I finally reached the solution. Thanks for all your support!!

7. Feb 20, 2014

I am working on the second part of this question and don't understand why the (7/16) is to the power of 3/2. Am I just missing something obvious?

Last edited: Feb 20, 2014
8. Feb 20, 2014

### Staff: Mentor

In order to be able to comment we'll have to see your work.