Electric Field due to Charged Disc

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Last updated on September 10th, 2019 at 03:22 am

The main objective of this article is to calculate the Electric Field due to Charged Disc. In our previous article, we have discussed the Electric Field due to Uniformly Charged Ring. We will use the final expression from that article as a starting point of our discussion. So if you are not familiar with, you can go back and start from there. Moreover, you can start with our main Electrostatics article.

Electric Field due to Charged Disc

Let us consider a uniformly charged disc of total charge Q. The center of the disc is located at point O. We have to calculate the electric field due to this uniformly charged disc at an axial point P. This is shown in the figure below. The distance between O and P is r. We also have shown the direction of the resultant electric field which is along the direction of \vec{OP} in the figure below.Electric Field due to Charged Disc

In order to calculate the electric field due to the whole disc, we will divide the disc into several concentric circles of varying radii and integrate the contributions from each circle. Let us draw one such circle of radius x. Then, we draw another concentric circle of radius x+dx. These two circles form a disc of thickness dx. This is shown as a red shaded region in the figure above. The area of this ring is 2\pi x dx and the corresponding charge contained in this disc is dq=\frac{Q}{\pi a^2}(2\pi x dx)= \frac{2Qx dx}{a^2}. Here, we have divided the total charge (Q) by the area of the disc (2\pi a^2) to get the surface charge density and then have multiplied the charge density by the area of the elemental ring to get the charge contained in the ring.

As discussed in our previous article the electric field due to this elemental ring is

    \[dE=\frac{dq r}{4\pi\epsilon_0 (r^2+x^2)^{3/2}} \]

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    \[=\frac{\left(\frac{2Qxdx}{a^2}\right)r}{4\pi\epsilon_0(r^2+x^2)^{3/2}} \]

    \[=\frac{Qr}{a^2}\frac{xdx}{2\pi\epsilon_0 (r^2+x^2)^{3/2}} \]

We will integrate over the full disc to calculate the electric field due to this uniform disc. Therefore, we will vary x from 0 to a. Thus

    \[E=\int_{0}^{a}{\frac{Qr}{2\pi\epsilon_0 a^2}\frac{xdx}{(r^2+x^2)^{3/2}}} \]

    \[=\frac{Q}{2\pi\epsilon_0 a^2}\left[1-\frac{r}{\sqrt{r^2+a^2}}\right]\]

    \[=\frac{Q}{2\pi\epsilon_0 a^2}(1-cos(\theta))\]

Or

    \[\vec{E}=\frac{Q}{2\pi\epsilon_0a^2}(1-cos(\theta))\hat{r}\]

electric field due charged disc as a function of distance

Here we have plotted the electric field due to a uniformly charged disc at a point on its axis as a function of distance from the center of the disc.

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