Charges on an equilateral triangle

AI Thread Summary
The discussion revolves around calculating the forces and electric fields acting on a charge q located at the top corner of an equilateral triangle, with charges Q at the other two corners. For part (a), the net force on charge q is directed vertically, but the initial calculations for the force were incorrect due to improper handling of vector components. In part (b), the electric field equations for q and Q were also miscalculated, prompting suggestions to use trigonometric functions or the Pythagorean theorem for accurate results. Participants emphasized the importance of breaking down the forces into their respective components to solve the problem correctly. Overall, the discussion highlights the need for careful analysis of forces in a triangular configuration.
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Homework Statement


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?


Homework Equations



F=(1/4πε) * (q1q2/r2)

E = (1/4πε) * (q/r2)

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?

Thanks a lot in advance.
 
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4real4sure said:

Homework Statement


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?


Homework Equations



F=(1/4πε) * (q1q2/r2)

E = (1/4πε) * (q/r2)

The Attempt at a Solution



(a) F = qQ / (4πεa2) iy
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.
(b) Eq = q / (4πε (3/16 a2)) iy

and EQ = Q / (4πε (7/16 a2)) iy
Those electric field calculations are not quite right either.
 
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?
 
Did you begin with a diagram of the triangle, labeling all sides in cluding the altitude?
attachment.php?attachmentid=45937&stc=1&d=1333748959.gif

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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4real4sure said:
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?
Let's start with the first part (a).

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?
 
I tried and I finally reached the solution. Thanks for all your support!
 
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:
lystrade said:
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?

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