- #1
bjornert
- 5
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Ok, so here's the question. I've pretty much got it except I'm having trouble with one part of constructing the integral. The problem is...
An insulator which lies between the positions -d[tex]\hat{z}[/tex] and d[tex]\hat{z}[/tex] has a nonuniform linear charge density [tex]\lambda[/tex] = [tex]\lambda_{o}[/tex][tex]\frac{z}{d}[/tex] . Find the force on a charge 'q' located at 3d[tex]\hat{z}[/tex].
[tex]\int[/tex] d[tex]\vec{E}[/tex] = [tex]\int[/tex][tex]\frac{k dq}{r^{2}}[/tex][tex]\hat{z}[/tex]
dq = [tex]\lambda[/tex]dz
[tex]\lambda[/tex] = [tex]\lambda_{o}[/tex][tex]\frac{z}{d}[/tex]
What I ended up getting when constructing the integral is...
[tex]\int[/tex][tex]^{ d}_{-d}[/tex]k[tex]\frac{\lambda_{0} z dz}{d r^{2}}[/tex][tex]\hat{z}[/tex]
Now I'm having trouble with the r[tex]^{2}[/tex] part in the denominator...I know it should be something like...
(r [tex]\pm[/tex] something)[tex]^{2}[/tex]
...but I'm just not sure what it is...any help would be greatly appreciated.
Thanks!
An insulator which lies between the positions -d[tex]\hat{z}[/tex] and d[tex]\hat{z}[/tex] has a nonuniform linear charge density [tex]\lambda[/tex] = [tex]\lambda_{o}[/tex][tex]\frac{z}{d}[/tex] . Find the force on a charge 'q' located at 3d[tex]\hat{z}[/tex].
Homework Equations
[tex]\int[/tex] d[tex]\vec{E}[/tex] = [tex]\int[/tex][tex]\frac{k dq}{r^{2}}[/tex][tex]\hat{z}[/tex]
dq = [tex]\lambda[/tex]dz
[tex]\lambda[/tex] = [tex]\lambda_{o}[/tex][tex]\frac{z}{d}[/tex]
The Attempt at a Solution
What I ended up getting when constructing the integral is...
[tex]\int[/tex][tex]^{ d}_{-d}[/tex]k[tex]\frac{\lambda_{0} z dz}{d r^{2}}[/tex][tex]\hat{z}[/tex]
Now I'm having trouble with the r[tex]^{2}[/tex] part in the denominator...I know it should be something like...
(r [tex]\pm[/tex] something)[tex]^{2}[/tex]
...but I'm just not sure what it is...any help would be greatly appreciated.
Thanks!