Electric Potential on z-axis due to circular disk

[sandy.bridge]

Homework Statement

Hey guys, I am a little bit confused how my textbook arrived at the answer it did for this particular question. The disk is in the xy-plane with radius b and uniform charge density. They go

$$V=\frac{\rho_S}{4πε}\int_0 ^{2π}\int_0 ^b\frac{r'}{(r'^2+z^2)^{1/2}}dr'd\phi'=\frac{ρ_S}{2ε}[(z^2+b^2)^{1/2}-|z|]$$

Where does the|z| come from? When I evaluate the integral, I do not get that.
I get
$$\frac{ρ_S}{2ε}[(z^2+b^2)^{1/2}]$$

 

[collinsmark]

Homework Statement

Hey guys, I am a little bit confused how my textbook arrived at the answer it did for this particular question. The disk is in the xy-plane with radius b and uniform charge density. They go

$$V=\frac{\rho_S}{4πε}\int_0 ^{2π}\int_0 ^b\frac{r'}{(r'^2+z^2)^{1/2}}dr'd\phi'=\frac{ρ_S}{2ε}[(z^2+b^2)^{1/2}-|z|]$$

Where does the|z| come from? When I evaluate the integral, I do not get that.
I get
$$\frac{ρ_S}{2ε}[(z^2+b^2)^{1/2}]$$

Try to evaluate your limits again.

Here is a hint: Evaluating,

$$\sqrt{z^2 + r^2} \ \ \bigg|_{r=0}^b$$
is not just $\sqrt{z^2 + b^2}$.

 

[sandy.bridge]

! Thank you!