Charges on an equilateral triangle

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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 alot in advance.
 

Answers and Replies

  • #2
collinsmark
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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
No, that's not quite right.

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

But the [itex] \hat \imath_y [/itex] 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.
 
  • #3
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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
gneill
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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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  • #5
collinsmark
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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

[tex] \vec F = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{r_2} \hat \imath_r. [/tex]

where [itex] \hat \imath_r [/itex] points in the direction from the particular Q to q.

But the [itex] \hat \imath_r [/itex] for one charge is different from the [itex] \hat \imath_r [/itex] 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 [itex] \hat \imath_r [/itex] into its [itex] \hat \imath_x [/itex] and [itex] \hat \imath_y [/itex] components.

That might involve something like

[tex] F_x = F\cos \theta [/tex]
[tex] F_y = F \sin \theta [/tex]
(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
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I tried and I finally reached the solution. Thanks for all your support!!
 
  • #7
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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:
  • #8
gneill
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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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