 #1
zenterix
 61
 20
 Homework Statement:
 A disc of radius ##R## is uniformly charged with total charge ##Q>0##. Determine the direction and magnitude of the electric field at a point P lying a distance ##x>0## from the center of the disc along the axis of symmetry of the disc.
 Relevant Equations:

I know how to solve this problem. The electric field vector points in only one direction, so there is really only one integral to do. However, I attempted to do the full integral, expecting the other two directions to cancel out, but my integrals don't seem to work out to the correct values. This is thus a question about integration.
Here I set up the integral for the electric field due to the disk at a point ##P##:
$$d\vec{E}_p=\frac{k_edq}{r_{dq,p}^2}\hat{r}_{dq,p}$$
$$dq=2\pi r dr \frac{Q}{\pi R^2}=\frac{2Qrdr}{R^2}$$
$$\vec{r}_{dq,p}=\vec{r}_{o,p}\vec{r}_{o,dq}=x\hat{k}r\hat{r}$$
$$r_{dq,p}=\sqrt{x^2+r^2}$$
$$\implies \vec{E}_p=\frac{2k_eQrdr}{R^2(x^2+r^2)^{3/2})}(x\hat{k}r\hat{r})$$
$$\vec{E}_p=\frac{2k_eQ}{R^2}\left[\int_0^R \frac{rx}{(x^2+r^2)^{3/2}}dr\hat{k}\int_0^R \frac{r^2}{(x^2+r^2)^{3/2}}dr\hat{r}\right]$$
I am interested in particular in the second integral, in the ##\hat{r}## direction.
Here is my depiction of the problem:
As far as I can tell, due to the symmetry of the problem, this integral should be zero.
$$\int_0^R \frac{r^2}{(x^2+r^2)^{3/2}}dr\hat{r}$$
I don't believe I need to integrate anything inside ##\hat{r}##; it depends on an angle relative to some axis but not on ##r##.
$$r=x\tan{\alpha}$$
$$dr=x \sec^2{\alpha}d\alpha$$
$$(x^2+r^2)^{3/2}=(x^2(1+\frac{r^2}{x^2}))^{3/2}=x^3(1+\tan^2{\alpha})^{3/2}=x^3\sec^3{\alpha}$$
So the integral becomes
$$\int \frac{x^2\tan^2{\alpha} x \sec^2{\alpha}}{x^3\sec^3{\alpha}}d\alpha$$
$$=\int \frac{\tan^2{\alpha}}{\sec{\alpha}}d\alpha$$
$$=\int \frac{\sin^2{\alpha}}{\cos{\alpha}}d\alpha$$
$$=\int \frac{1}{\cos{\alpha}}d\alpha=\int_0^R \cos{\alpha}d\alpha$$
$$=(\ln\sec{\alpha}+\tan{\alpha}  \sin{\alpha})$$
Note that
$$r=x\tan{\alpha}$$
$$\implies \sin{\alpha}=\frac{r}{(x^2+r^2)^{1/2}}$$
$$\implies \cos{\alpha}=\frac{x}{(x^2+r^2)^{1/2}}$$
The integral result in terms of ##r## is then
$$(\ln\frac{(x^2+r^2)^{1/2}}{x}+\frac{r}{x}\frac{r}{(x^2+r^2)^{1/2}}) \big_0^R$$
Which is not zero.
I don't know if I messed up setting up the integral, or in the integration itself.
Here is my depiction of the problem:
As far as I can tell, due to the symmetry of the problem, this integral should be zero.
$$\int_0^R \frac{r^2}{(x^2+r^2)^{3/2}}dr\hat{r}$$
I don't believe I need to integrate anything inside ##\hat{r}##; it depends on an angle relative to some axis but not on ##r##.
$$r=x\tan{\alpha}$$
$$dr=x \sec^2{\alpha}d\alpha$$
$$(x^2+r^2)^{3/2}=(x^2(1+\frac{r^2}{x^2}))^{3/2}=x^3(1+\tan^2{\alpha})^{3/2}=x^3\sec^3{\alpha}$$
So the integral becomes
$$\int \frac{x^2\tan^2{\alpha} x \sec^2{\alpha}}{x^3\sec^3{\alpha}}d\alpha$$
$$=\int \frac{\tan^2{\alpha}}{\sec{\alpha}}d\alpha$$
$$=\int \frac{\sin^2{\alpha}}{\cos{\alpha}}d\alpha$$
$$=\int \frac{1}{\cos{\alpha}}d\alpha=\int_0^R \cos{\alpha}d\alpha$$
$$=(\ln\sec{\alpha}+\tan{\alpha}  \sin{\alpha})$$
Note that
$$r=x\tan{\alpha}$$
$$\implies \sin{\alpha}=\frac{r}{(x^2+r^2)^{1/2}}$$
$$\implies \cos{\alpha}=\frac{x}{(x^2+r^2)^{1/2}}$$
The integral result in terms of ##r## is then
$$(\ln\frac{(x^2+r^2)^{1/2}}{x}+\frac{r}{x}\frac{r}{(x^2+r^2)^{1/2}}) \big_0^R$$
Which is not zero.
I don't know if I messed up setting up the integral, or in the integration itself.