# Electric Flux Through a Circular Disc due to a Point Charge

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.

• Have you computer the two integrals to see if they are the same? That would be the quickest way to tell. Apr 23, 2022 at 21:08

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.