Use integration to find the total force on a point HELP

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http://capa.mcgill.ca/res/mcgill/dcmcgill/oldproblems/newerlib/Graphics/Gtype51/prob19a.gif[/URL]

1. A charge Q = 8.10 ×10-4 C is distributed uniformly along a rod of length 2L, extending from y = -11.4 cm to y = +11.4 cm, as shown in the diagram 'on your assignment above. A charge q = 1.80 ×10-6 C, and the same sign as Q, is placed at (D,0), where D = 17.5 cm.
Use integration to compute the total force on q in the x-direction




It is already proven that :
The magnitude of the force on charge q due to the small segment dy is
dF=(kqQ/2Lr²)dy.



3. Given what's already proven that for a small dy:
dF = (kqQ/2Lr²)dy
so i integrated both sides and i got:
∫dF = k*q*Q / 2L ∫ 1/r^2 dy

since r^2 = D^2+y^2

therefore the function becomes:
∫dF = k*q*Q / 2L ∫ 1/D^2+y^2 dy

now what should i do next? and the equation i derived, was it right at the first place?


THANKS!

 

[Andrew Mason]

http://capa.mcgill.ca/res/mcgill/dcmcgill/oldproblems/newerlib/Graphics/Gtype51/prob19a.gif[/URL]

1. A charge Q = 8.10 ×10-4 C is distributed uniformly along a rod of length 2L, extending from y = -11.4 cm to y = +11.4 cm, as shown in the diagram 'on your assignment above. A charge q = 1.80 ×10-6 C, and the same sign as Q, is placed at (D,0), where D = 17.5 cm.
Use integration to compute the total force on q in the x-direction




It is already proven that :
The magnitude of the force on charge q due to the small segment dy is
dF=(kqQ/2Lr²)dy.

That may be the magnitude of the force. But what is the magnitude of the x component of that force? (You can see by the symmetry, that the y components all sum to 0 so the integral of the x component gives you the total force).

AM

 

[lmlgrey]

oh, i see...
so now since only the x-component is considered then
cos theta = D/r
and the function becomes:
dF=(kqQ/2Lr^2)dy * D/ r ... is that correct?

then further integrating the above gives:

F= k*q*Q*D^2/2L* ∫1/(D^2+y^2)^-2/3 8 dy? -- did i do this step correct??

 

[Andrew Mason]

oh, i see...
so now since only the x-component is considered then
cos theta = D/r
and the function becomes:
dF=(kqQ/2Lr^2)dy * D/ r ... is that correct?

then further integrating the above gives:

F= k*q*Q*D^2/2L* ∫1/(D^2+y^2)^-2/3 8 dy? -- did i do this step correct??

Not quite.

$$F = \frac{kqQD}{2L}\int_{-L}^{L}\frac{1}{r^3} dy = \frac{kqQD}{2L}\int_{-L}^{L}\frac{1}{(\sqrt{D^2 + y^2})^3} dy =\frac{kqQD}{2L}\int_{-L}^{L}(D^2 + y^2)^{-\frac{3}{2}}dy$$

Good luck working out that integral

AM

 

[lmlgrey]

thanks, i solved it :)

 

[adchang]

http://img232.imageshack.us/img232/4277/hardsk4.th.jpg

can anyone show me how to do this integral manually?