Electric potential - a disc

Question

I have a small question. The electric field exerted by a disc with radius $R$ (charge density $\sigma$) on the $xy$ plane on a point $z$ on the $z$-axis is:

$$\vec{E}=\frac{\sigma}{2\epsilon_0}[1-\frac{z}{\sqrt {z^2+R^2}}]\hat{z}.$$

Now, if I want to calculate the electric potential at a point $(0,0,D)$ on the $z$-axis, I need to calculate:

$$\phi=-\int_\infty ^D \frac{\sigma}{2\epsilon_0}[1-\frac{z}{\sqrt {z^2+R^2}}]dz.$$

Now the first integrand gives $-\int_\infty ^D dz$ which is $\infty-D$.

My question is then, what makes the integral converge? is it the second integrand? two orders of "the same infinity"? Isn't it an approximation then?

Answer 1

You can expand the term $\frac{z}{\sqrt{z^2+R^2}}$ using binomial expansion as $$1 - \frac{R^2}{2z^2} + \ldots$$ Thus, the first term, which is also $1$, cancels with the $1$ at the beginning and the rest of the terms converge.

For large $z$, we can neglect higher terms and the $R^2/2z^2$ term becomes dominant. Integrating it gives a result proportional to $1/z$, which is exactly as expected because the potential behaves like a point charge when $z$ is very large.

Answer 2

Yes, the integral does converge because the infinities deriving from the two terms cancel out, being of the same order ($\sqrt{z^2+R^2}$ behaves like $z$ when $z \rightarrow \infty$).

However, that does not mean that you cannot obtain the exact expression. In fact you can solve the integral: $$ \Phi = - \frac{\sigma}{2\epsilon_0} \int_{+\infty}^D dz \left[1- \frac{z}{\sqrt{z^2+R^2}}\right] = - \frac{\sigma}{2\epsilon_0} \left[z - \sqrt{z^2+R^2}\right]_{+\infty}^D $$ and you will get a $[+\infty - \infty]$ indeterminate form, that you can solve by rationalising:

$$ \Phi = -{\sigma \over 2\epsilon_0} \left[(z - \sqrt{z^2+R^2}) \cdot \frac{z + \sqrt{z^2+R^2}}{z + \sqrt{z^2+R^2}}\right]_{+\infty}^D $$

Finally you will get:

$$ \Phi = +{\sigma \over 2\epsilon_0} \left[ {R^2 \over z + \sqrt{z^2+R^2}} \right]_{+\infty}^D = \frac{\sigma}{2\epsilon_0} \cdot {R^2 \over D + \sqrt{D^2+R^2}} $$ Note that no approximation has been made.

PS: this result, as well as the expression of $\vec{E}$ above, holds for $D>0$: however, if $D<0$ we can simply substitute $|D|$ to $D$ in the expression of $\Phi$, as the problem is symmetric respect to the $xy$ plane.

Answer 3

You can understand the calculation better if you see this equation plots. For the part $\frac{z}{\sqrt{z^2+R^2}}$ it goes to 1 if z approaches to infinity. So the area under this curve (Which we know it's $\sqrt{z^2+R^2}$ ) goes to infinity. Also the area under 1 (that is $z$) is infinity.

So the equation is $[\infty - \infty] $ which is indeterminate but it can be solved.

Answer 4

To calculate electric potential at a point from this electric field, we need to do indefinite integration of the electric field, and then evaluate the result by substituting $z=D$ :
$V(z)=-\frac{\sigma }{2 \epsilon _0}\int \left[1-\frac{z}{\sqrt{R^2+z^2}}\right]\text{dz}=\frac{\sigma }{2 \epsilon _0}\left(\sqrt{R^2+z^2}-z\right)$
Then $V(D)=\frac{\sigma }{2 \epsilon _0}\left(\sqrt{R^2+D^2}-D\right)$
It’s easier to calculate directly the electric potential though, as electric potential is a scalar quantity.