Electric Flux Through a Circular Disc due to a Point Charge

Question

I am having trouble understanding the proposed method for finding the electric flux through a disc of radius $a$ given by a point charge at distance $z_0$. $$ \int \vec{E} \cdot \hat{n}da = \int_0^{tan^{-1}a/z_0} \frac{q}{a^2+z_0^2} 2\pi (a^2 +z_0^2) \sin{\theta} d\theta $$ I have seen other solutions that describe the same problem but in a different manner. In this case, the flux is found using the differential ring element $2\pi xdx$ and the value of $\cos(\theta)$. $$ \int \vec{E} \cdot \hat{n}da = \int_0^{a} \frac{q}{x^2+z_0^2} 2\pi x \frac{z_0}{\sqrt{z_0^2 + x^2}} dx $$ Am I wrong to say that the problems described are the same? Or is it just another way to getting the same result? Thanks.

Answer 1

Im assuming that the charge is located perpendicular to the disk with distance $z_0$ directly above its center. Lets do the second integral first.

The orthogonal component of the electric field that passes through the disk is given by $$ \frac{q z_0}{\sqrt{x^2 + z_0^2}^3} $$ Now, as you said, we will integrate over the whole surface of the disk by using increasing anulli of width $dx$ $$ \int_0^{a}\int_0^{2 \pi}{\frac{q z_0}{\sqrt{x^2 + z_0^2}^3} x d\varphi dx} = 2 \pi q z_0 \int_0^{a}{\frac{x}{\sqrt{x^2 + z_0^2}^3} dx} = 2 \pi q \bigg(1-\frac{z_0}{\sqrt{z_0^2+a^2}}\bigg) $$

Now lets do the first integral. Ill call $\tan^{-1}(a/z_0) =: \Theta$ for now. Ill also cancel out the $(a^2+z_0^2)$ $$ 2 \pi q \int_0^\Theta{\sin\theta d\theta} = 2 \pi q (\cos(0) - \cos(\tan^{-1}(a/z_0))) = 2 \pi q \bigg(1 - \frac{z_0}{\sqrt{z_0^2+a^2}}\bigg) $$ So yes, the integrals are the same. Although, the last formula (your first one) does not seem very intuitive to me. If you really want to integrate over the angle i would suggest substituting properly with $x = z_0 \tan(\theta) \Rightarrow dx = \frac{z_0 d\theta}{\cos^2(\theta)}$. Lets start with the intuitive formula $$ 2 \pi q z_0 \int_0^{a}{\frac{x}{\sqrt{x^2 + z_0^2}^3} dx} = 2 \pi q z_0 \int_0^{\Theta}{\frac{z_0 \tan(\theta)}{\sqrt{z_0^2 \tan^2(\theta) + z_0^2}^3} \frac{z_0}{\cos^2(\theta)} d\theta} $$ Canceling the $z_0$'s and using $\sin^2+\cos^2 = 1$ in the denominator gives $$ 2 \pi q \int_0^{\Theta}{\cos(\theta) \tan(\theta) d\theta} = 2 \pi q \int_0^\Theta{\sin\theta d\theta} $$ Which will again yield the same result.