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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...

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