Electric Potential due to Circular Disc

Electric Potential due to Circular Disc

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Last updated on August 29th, 2019 at 04:04 pm

In this article, we will discuss the Electric Potential due to Circular Disc. However, you can browse our previous article, Electric Potential due to a Charged Ring which is the prerequisite of this discussion. Also, you can also check the article Electrostatic Potential to start from scratch. We define the electrostatic potential as the work done to move a unit positive charge from a point of reference, generally chosen to be infinity to the field of influence. However, to calculate the electric potential here, we will start from our previous result and will calculate the potential due to an elemental circular ring. Then we will integrate over the whole disc to get the total potential at a point on its axis. No surprise that this methodology is very similar to the calculation of Gravitational Potential due to a Circular Disc. Also, notice that this is an example of electrostatic potential due to a uniform surface charge distribution.

Electric Potential due to Circular Disc

Here we will calculate the electrostatic potential due to a charged circular disc of radius R at an axial point P. The uniformly charged disc is centered at O. The distance of the test point P from the center O is Z. We will consider a differential charged ring element with a radius of r and thickness dr. Let the charge contain within this thick circular ring is dq. The potential at point P due to this charged ring is given by  dV(z)=\frac{dq}{4 \pi \epsilon_0 \sqrt{r^2+z^2}}}. Here dq is the charge contained in the blue shaded region.

electric potential due to charged circular disc

The area of the blue shaded region is  2\pi r dr. Let Q be the total charge of the disc. Then the surface charge density is  \sigma= \frac{Q}{\pi R^2} and thus the charge contained within this elemental area is  \sigma 2\pi r dr = \frac{2Qr dr}{R^2}. The electric potential at point P due to this ring element is

    \[  dV(z)= \frac{2Qr dr}{4 \pi \epsilon_0 R^2 \sqrt{r^2+z^2}} \]

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    \[  V(z)= \frac{Q}{2 \pi \epsilon_0 R^2} \int_0^R \left \frac{r dr}{\sqrt{r^2+z^2}}  \right \]

Thus the total electric potential due to this charged disc is

    \[  = \frac{Q}{2 \pi \epsilon_0 R^2} \left [ \sqrt{r^2+z^2} \right]_0^R \]

    \[  \boxed{ V(z)= \frac{Q}{2 \pi \epsilon_0 R^2} \left [ \sqrt{R^2+z^2} - z \right] } \]

This is the final expression for the electric potential due to a charged circular disc at a point P on its axis. The electrostatic potential is a scalar quantity which makes this calculation very easy.

At a very large distance i.e. Z>>R, the approximate potential is

    \[  V(z) \cong \frac{Qz}{2 \pi \epsilon_0 R^2}\left[ (1+\frac{R^2}{2z^2})-1\right] \]

    \[  \cong \frac{Q}{4\pi \epsilon_0 z} \]

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