# Use integration to find the total force on a point HELP!

1. Jan 13, 2009

### lmlgrey

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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 whats 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!

Last edited by a moderator: Apr 24, 2017 at 10:16 AM
2. Jan 13, 2009

### Andrew Mason

Last edited by a moderator: Apr 24, 2017 at 10:16 AM
3. Jan 13, 2009

### 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??

4. Jan 14, 2009

### Andrew Mason

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

5. Jan 14, 2009

### lmlgrey

thanks, i solved it :)

6. Jan 15, 2009