Finding the force due to uniform charged rod

  • Thread starter mattbonner
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  • #1
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Homework Statement



A charge Q = 4.35 ×10−4 C is distributed uniformly along a rod of length 2L, extending from y = -21.1 cm to y = +21.1 cm, as shown in the diagram 'on your assignment above. A charge q = 6.05 ×10−6 C, and the same sign as Q, is placed at (D,0), where D = 74.5 cm.

Use integration to compute the total force on q in the x-direction.

Homework Equations


[tex]E = k \int \frac{dq}{r^2}[/tex]
F = qE

The Attempt at a Solution



I've seen the answer, and I think I'm missing a step because when I do it it always turns out different

[tex]E = k \int_{-.211}^{.211} \frac{dq}{(\sqrt{D^2 + y^2})^2} [/tex] * [tex] \frac{D}{\sqrt{y^2 + D^2}}[/tex]


when I simplify this I get,

[tex] F = kqQD \int_{-L}^{L}(D^2 + y^2)^{-\frac{3}{2}}dy [/tex]

however, the correct answer is

[tex] F = \frac{kqQD}{2L}\int_{-L}^{L}(D^2 + y^2)^{-\frac{3}{2}}dy[/tex]

and I'm confused as to where the 2L comes from...
 
Last edited:

Answers and Replies

  • #2
Nabeshin
Science Advisor
2,205
16
I've seen the answer, and I think I'm missing a step because when I do it it always turns out different

[tex]E = k \int_{-.211}^{.211} \frac{dq}{(\sqrt{D^2 + y^2})^2} [/tex] * [tex] \frac{D}{\sqrt{y^2 + D^2}}[/tex]


when I simplify this I get,

[tex] F = kqQD \int_{-L}^{L}(D^2 + y^2)^{-\frac{3}{2}}dy [/tex]
I don't know why you're multiplying by what you're multiplying in your first step: you haven't clearly shown your approach (and this is why, I presume, you're missing a small factor).

You correctly substituted in at this step:
[tex]E = k \int_{-.211}^{.211} \frac{dq}{(\sqrt{D^2 + y^2})^2} [/tex]
replacing r with the Pythagorean distance, but could you explain why you multiplied by that factor? (this should help you realize your mistake)
 
Last edited:
  • #3
14
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Due to symmetry, I multiplied by cosine, which I think is D/r
 
  • #4
Nabeshin
Science Advisor
2,205
16
Ok, to find the X component of the electric field, if I'm following you correctly.

So you have:
[tex]E_{x}=kD\int_{-L}^{L}\frac{dq}{(y^{2}+D^{2})^{\frac{3}{2}}}=kD\int_{-L}^{L}(y^{2}+D^{2})^{-\frac{3}{2}}dq[/tex]

Which is just simplifying the multiplication.

But you still are integrating with respect to q, without any q's in the equation! So you need to find some way to transform dq into dy (what you want to be integrating with respect to).
 
  • #5
14
0
Ok, to find the X component of the electric field, if I'm following you correctly.

So you have:
[tex]E_{x}=kD\int_{-L}^{L}\frac{dq}{(y^{2}+D^{2})^{\frac{3}{2}}}=kD\int_{-L}^{L}(y^{2}+D^{2})^{-\frac{3}{2}}dq[/tex]

Which is just simplifying the multiplication.

But you still are integrating with respect to q, without any q's in the equation! So you need to find some way to transform dq into dy (what you want to be integrating with respect to).
ooh I think I'm starting to see it:
dq = [tex]\lambda dy[/tex]
and
[tex]\lambda = \frac{Q}{2L}[/tex]

and then [tex]\lambda[/tex] can be brought out of the integral and that's where the 2L comes from!

Thank you so much! This was bugging me for quite a while.
 

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