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Positive charge distributed uniformly along y axis

by nateastle
Tags: axis, charge, distributed, positive, uniformly
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nateastle
#1
Aug28-06, 09:29 PM
P: 15
I have a physics question that states:
An amount of positive charge is distributed uniformly along the positive y-axis between y=o and y=a. A negative point charge -q lies on teh positive x=axis a distance r from the origin. Derive the x and y compontes of the force that the charge distribution exerts on Q exerts on q.

I have figured the y force to be: (Qqk/a)[(q/x)-(1/(a^2 +x^2)^1/2)] I did this by drawing out the graph and by doing an intgral from 0 to a on dfsin theta. Where theta is the angle where the line comes from the top of through q. I then used trig substitution to figure out what sin theta is. The part that I am stuck on is how do I solve for the force on the X axis.

Any help is much appreciated.
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Andrew Mason
#2
Aug29-06, 07:45 AM
Sci Advisor
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P: 6,671
Quote Quote by nateastle
I have a physics question that states:
An amount of positive charge is distributed uniformly along the positive y-axis between y=o and y=a. A negative point charge -q lies on teh positive x=axis a distance r from the origin. Derive the x and y compontes of the force that the charge distribution exerts on Q exerts on q.

I have figured the y force to be: (Qqk/a)[(q/x)-(1/(a^2 +x^2)^1/2)] I did this by drawing out the graph and by doing an intgral from 0 to a on dfsin theta. Where theta is the angle where the line comes from the top of through q. I then used trig substitution to figure out what sin theta is. The part that I am stuck on is how do I solve for the force on the X axis.
The force on q of a charge [itex]dQ = \frac{Q}{a}dy[/itex] is:

[tex]F = \frac{kq}{(r^2+y^2)}dQ = \frac{kqQ}{a(r^2+y^2)}dy[/tex]

So the components of the Coulomb force on q would be:

[tex]F_x = \frac{kqQ}{a}\int_0^a \frac{1}{y^2+r^2}cos\theta dy[/tex]

[tex]F_y = \frac{kqQ}{a}\int_0^a \frac{1}{y^2+r^2}sin\theta dy[/tex]

where [itex]sin\theta = y/\sqrt{y^2+r^2}[/itex] and [itex]cos\theta = r/\sqrt{y^2+r^2}[/itex]

Work out those integrals and you should get the right answer.

AM


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