Question:
The portion of the z-axis for which $-2<z<2$ m carries a nonuniform charge density of $z^2 + 1$ nC/m in free space. There are no other charges anywhere. Find $\vec{E}$ at
a) $P_{A}(1,0,0)$ and
b) $P_{B}(2,0,0)$.
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Attempt at Solution:
I am honestly not really sure how to start. In previous problems of this nature, the charge has been uniform, so the symmetry caused there to be no z-component of the electric field, but now, since there is no longer a uniform density, I am stuck. I tried working the problem using some of the methods I had learned for uniform distributions just to try to make some progress and this is what I came up with:
$d\vec{E} = \frac{x(z^2 + 1)dz}{4\pi{}\epsilon_0(z^2 + x)^{3/2}}$.
In the case of part a), $d\vec{E} = \frac{(1)(z^2 + 1)dz}{4\pi{}\epsilon_0(z^2 + (1)^2)^{3/2}} = \frac{dz}{4\pi{}\epsilon_0(z^2 + 1)^{1/2}}$.
Integrating this from -2 to 2, I got $\vec{E} = \frac{1}{4\pi{}\epsilon_0}\ln{(2 + \sqrt{5})} - \ln{(-2 + \sqrt{5})} = 2.595*10^{-10}$ V/M. According to my thought process, this is in the x-direction.
Now here is where I really started to have trouble... In the case of part b), I got $d\vec{E} = \frac{(z^2 + 1)dz}{2\pi{}\epsilon_0(z^2 + 4)^{3/2}}$. I don't think I can integrate this expression.
Can anyone tell me what I am doing wrong?
Well we can surely integrate this expression $$ \int\frac{(z^2+1)dz}{(z^2+4)^{3/2}}=\int\frac{(z^2+4)dz}{(z^2+4)^{3/2}}-\int\frac{3dz}{(z^2+4)^{3/2}}$$ $$=\int\frac{dz}{(z^2+4)^{1/2}}-\int\frac{3dz}{(z^2+4)^{3/2}}$$ $$=\int\frac{dz}{2(1+(z/2)^2)^{1/2}}-\int\frac{3dz}{8(1+(z/2)^2)^{3/2}}$$ now substitute $\frac{z}{2}=t$ and we get $$=\int\frac{dt}{(1+t^2)^{1/2}}-\int\frac{3dt}{4(1+t^2)^{3/2}}$$ Now we can solve the first term same as part (a)
For the second term put $t=\tan\theta$ and $dt=\sec^2\theta\; d\theta$ , and we're done.
Hope that helps.