Uniformly Charged Disk!

Write the differential equation for the electric field. Define the variables, let the differential area be a function of the radius and the angle, and finally perform the double integral.

The electric field for an infinitesimally small charged section of the disc is:

\begin{equation*}
\vec{E}=\frac{1}{4\pi\epsilon_0}\int_{\text{Disk}}\frac{dq}{r’^2}\,\hat{\textbf{r}}’,
\end{equation*}

where \(\hat{\textbf{r}}’ = -r\cos(\phi)\,\hat{\textbf{i}}-r\sin(\phi)\,\hat{\textbf{j}}+h\hat{\textbf{k}} \), and the charge \(dq = \sigma r dr d \phi\). The integral then becomes:

\begin{equation*}
\vec{E}=\frac{1}{4\pi\epsilon_0}\int_{0}^{R}\dfrac{\sigma r dr }{(\sqrt{r^2+h^2})^3}\int_{0}^{2\pi}d\phi\left(-r\cos(\phi)\,\hat{\textbf{i}}-r\sin(\phi)\,\hat{\textbf{j}}+h\hat{\textbf{k}}\right),
\end{equation*}

which can be done by substitution. The result after performing the integral is:

\begin{equation*}
\vec{E}=\frac{Qh}{2\pi\epsilon_0 R^2}\left(\frac{1}{\sqrt{h^2}}-\frac{1}{\sqrt{R^2+h^2}}\right)\,\hat{\textbf{k}},
\end{equation*}

which, with numerical values, becomes:

\begin{equation*}
\vec{E}\approx 1.15\times 10^{9}\,\text{N/C}\,\hat{\textbf{k}}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

We need to calculate the magnitude of the electric field at a distance \(h\) from the center of the disk. In order to solve this problem, we’ll write down the explicit integral expression for the electric field of a charge distribution. Given the symmetry of the disk, we’ll use cylindrical coordinates \((r,\phi,z)\), which are more convenient than Cartesian coordinates \((x,y,z)\).

First, let’s draw the situation presented in the problem and name our variables, as seen in figure 1.

Figure 1: We place the origin of our coordinate systems at the center of the disk. The usual Cartesian \(x\), \(y\), \(z\) is shown as well as the cylindrical coordinates \(r\), \(\varphi\) and \(z\). On the left, a 3D drawing of the disk is shown, indicating the direction of the electric field \(\vec{dE}\) produced by the charge differential \(dq\). On the top-right, we have the geometric relations between the variables \(h\), \(r\) and \(r’\) with the angle \(\theta\). On the bottom-right, we show the disk from above in the XY plane, where the cylindrical variables \(r\) and \(\varphi\) indicate the position of the charge differential \(dq\)

As seen in figure 1, the disk is placed in the XY plane, and the point where we must calculate the electric field is located along the Z-axis, at a distance \(h\) from the center of the disk.

Now, we can write the general equation for the electric field \(\vec{E}\),

\begin{equation}
\label{efield}
\vec{E}=\frac{1}{4\pi\epsilon_0}\int_{\text{Disk}}\frac{dq}{r’^2}\,\hat{\textbf{r}}’,
\end{equation}

where \(dq\) is the charge differential, \(\vec{r}\,’\) is a vector whose magnitude \(r’\) is the distance from this charge differential to the point where the electric field is calculated and \(\hat{\textbf{r}}’\) is the unitary vector in the direction from \(dq\) to the point where the field is calculated. The physical constant \(\epsilon_0\) is known as the permittivity of free space and takes the value \(\epsilon_0=8.854\times10^{-12}\,\text{F/m}\).

We choose our charge differential to be at the cylindrical coordinates \((r,\phi,0)\), as seen in the first figure, which is equivalent to the Cartesian coordinates \(x=r\cos(\phi)\) , \(y=r\sin(\phi)\) and \(z=0\). The Cartesian coordinates of the point where we want to calculate the electric field are \(x=0\), \(y=0\) and \(z=h\). To find the vector \(\vec{r}\), we must take the difference of the coordinates of the point where \(\vec{E}\) is being calculated and where \(dq\) is located, namely,

\begin{equation}
\label{vecr}
\vec{r}\,’=-r\cos(\phi)\,\hat{\textbf{i}}-r\sin(\phi)\,\hat{\textbf{j}}+h\hat{\textbf{k}}.
\end{equation}

Then, we can take the magnitude of vector \(\vec{r}\) to find \(r\) and obtain

\begin{equation}
r’=|\vec{r}\,’|,
\end{equation}

\begin{equation}
r’=\sqrt{(-r\cos(\phi))^2+(-r\sin(\phi))^2+h^2},
\end{equation}

\begin{equation}
r’=\sqrt{r^2(\cos^2(\phi)+\sin^2(\phi))+h^2},
\end{equation}

\begin{equation}
\label{magr}
r’=\sqrt{r^2+h^2},
\end{equation}

where, in the last line, we have used the trigonometric identity \(\cos^2(\phi)+\sin^2(\phi)=1\). The result from equation \eqref{magr} can also be seen in the first figure. The unitary vector \(\hat{\textbf{r}}’\) can be calculated according to the following expression

\begin{equation}
\hat{\textbf{r}}’=\frac{\vec{r}\,’}{r’},
\end{equation}

which, after using the results of equations \eqref{vecr} and \eqref{magr}, we get

\begin{equation}
\hat{\textbf{r}}’=\frac{-r\cos(\phi)\,\hat{\textbf{i}}-r\sin(\phi)\,\hat{\textbf{j}}+h\hat{\textbf{k}}}{\sqrt{r^2+h^2}},
\end{equation}

or equivalently

\begin{equation}
\label{unitr}
\hat{\textbf{r}}’=-\frac{r\cos(\phi)}{\sqrt{r^2+h^2}}\,\hat{\textbf{i}}-\frac{r\sin(\phi)}{\sqrt{r^2+h^2}}\,\hat{\textbf{j}}+\frac{h}{\sqrt{r^2+h^2}}\,\hat{\textbf{k}}.
\end{equation}

Using the explicit expressions for \(r’\) and \(\hat{\textbf{r}}’\) given in equations \eqref{magr} and \eqref{unitr} into equation \eqref{efield}, we obtain

\begin{equation}
\vec{E}=\frac{1}{4\pi\epsilon_0}\int_{\text{Disk}}\frac{dq}{(\sqrt{r^2+h^2})^2}\left(-\frac{r\cos(\phi)}{\sqrt{r^2+h^2}}\,\hat{\textbf{i}}-\frac{r\sin(\phi)}{\sqrt{r^2+h^2}}\,\hat{\textbf{j}}+\frac{h}{\sqrt{r^2+h^2}}\,\hat{\textbf{k}}\right),
\end{equation}

which, after factorization of the term \(\sqrt{r^2+h^2}\), we get

\begin{equation}
\label{efield2}
\vec{E}=\frac{1}{4\pi\epsilon_0}\int_{\text{Disk}}\dfrac{dq}{(\sqrt{r^2+h^2})^3}\left(-r\cos(\phi)\,\hat{\textbf{i}}-r\sin(\phi)\,\hat{\textbf{j}}+h\hat{\textbf{k}}\right).
\end{equation}

Because the charge is uniformly distributed, we can define a charge density \(\sigma\) by dividing the total charge \(Q\) by the total area of the disk of radius \(R\); explicitly,

\begin{equation}
\label{sigma}
\sigma=\frac{Q}{\pi R^2}.
\end{equation}

We can use this charge density \(\sigma\) to write \(dq\) in terms of the area differential \(dA\) as

\begin{equation}
\label{dq1}
dq=\sigma dA.
\end{equation}

In Cartesian coordinates, the area differential will be \(dA=dxdy\), but because we are working with cylindrical coordinates, the area differential on the XY plane is

\begin{equation}
\label{areacyl}
dA=rdrd\phi.
\end{equation}

Using the result of equation \eqref{areacyl} in equation \eqref{dq1}, we get

\begin{equation}
\label{dq2}
dq=\sigma r dr d\phi.
\end{equation}

We can now use the expression for \(dq\) given in equation \eqref{dq2} in equation \eqref{efield2} to write

\begin{equation}
\label{efield3}
\vec{E}=\frac{1}{4\pi\epsilon_0}\int_{\text{Disk}}\dfrac{\sigma r dr d\phi }{(\sqrt{r^2+h^2})^3}\left(-r\cos(\phi)\,\hat{\textbf{i}}-r\sin(\phi)\,\hat{\textbf{j}}+h\hat{\textbf{k}}\right).
\end{equation}

Now, we can see that we have to perform two integrals, one of for the variable \(r\) and one for the variable \(\phi\). For angle \(\phi\) we have a complete circle, so \(0\leq\phi\leq2\pi\).  For the \(r\) integral, the limits go from the center of the disk \(r=0\) to the total radius \(r=R\). We can then write the expression for \(\vec{E}\) in equation \eqref{efield3} as

\begin{equation}
\label{efield4}
\vec{E}=\frac{1}{4\pi\epsilon_0}\int_{0}^{R}\dfrac{\sigma r dr }{(\sqrt{r^2+h^2})^3}\int_{0}^{2\pi}d\phi\left(-r\cos(\phi)\,\hat{\textbf{i}}-r\sin(\phi)\,\hat{\textbf{j}}+h\hat{\textbf{k}}\right),
\end{equation}

where we have separated the terms that involve the variable \(\phi\) from those that do not. We can perform the \(\phi\) integral as follows using the distributive law

\begin{equation}
-r\int_{0}^{2\pi}\cos(\phi)d\phi\,\hat{\textbf{i}}-r\int_{0}^{2\pi}\sin(\phi)d\phi\,\hat{\textbf{j}}+h\int_{0}^{2\pi}d\phi\,\hat{\textbf{k}}.
\end{equation}

Performing each integral separately, we get

\begin{equation}
\int_{0}^{2\pi}d\phi\left(-r\cos(\phi)\,\hat{\textbf{i}}-r\sin(\phi)\,\hat{\textbf{j}}+h\hat{\textbf{k}}\right)=-r\sin(\phi)\Big|_{0}^{2\pi}\,\hat{\textbf{i}}+r\cos(\phi)\Big|_{0}^{2\pi}\,\hat{\textbf{j}}+h\phi\Big|_{0}^{2\pi}\,\hat{\textbf{k}},
\end{equation}

the terms in the \(\hat{\textbf{i}}\) and \(\hat{\textbf{j}}\) directions are zero, which we could expect from symmetry, and the result of the integral is

\begin{equation}
\label{intphi}
\int_{0}^{2\pi}d\phi\left(-r\cos(\phi)\,\hat{\textbf{i}}-r\sin(\phi)\,\hat{\textbf{j}}+h\hat{\textbf{k}}\right)=2\pi h\,\hat{\textbf{k}}.
\end{equation}

Using the result of equation \eqref{intphi} into equation \eqref{efield4}, we get the expression

\begin{equation}
\vec{E}=\frac{1}{4\pi\epsilon_0}\int_{0}^{R}\dfrac{2\pi h\sigma r dr }{(\sqrt{r^2+h^2})^3}\,\hat{\textbf{k}},
\end{equation}

which, after taking the constant terms out of the integral and simplifying, is

\begin{equation}
\label{efield5}
\vec{E}=\frac{\sigma h}{2\epsilon_0}\int_{0}^{R}\frac{rdr}{(r^2+h^2)^{3/2}}\,\hat{\textbf{k}},
\end{equation}

where we have written \((\sqrt{r^2+h^2})^3=(r^2+h^2)^{3/2}\)

This integral must be made using substitution. We choose the substitution \(u=r^2+h^2\), then \(du=2rdr\) or equivalently \(\frac{du}{2}=rdr\). Using these results in equation \eqref{efield5}, we get

\begin{equation}
\label{efield6}
\vec{E}=\frac{\sigma h}{2\epsilon_0}\int_{h^2}^{R^2+h^2}\frac{du}{2u^{3/2}}\,\hat{\textbf{k}}.
\end{equation}

Performing the integral, we obtain

\begin{equation}
\vec{E}=\frac{\sigma h}{2\epsilon_0}\left(-\frac{1}{u^{1/2}}\right)\Big|_{h^2}^{R^2+h^2}\,\hat{\textbf{k}},
\end{equation}

\begin{equation}
\vec{E}=\frac{\sigma h}{2\epsilon_0}\left(-\frac{1}{\sqrt{R^2+h^2}}+\frac{1}{\sqrt{h^2}}\right)\,\hat{\textbf{k}}.
\end{equation}

Using the explicit expression for \(\sigma\) given in equation \eqref{sigma}, we get

\begin{equation}
\vec{E}=\frac{Qh}{2\pi\epsilon_0 R^2}\left(\frac{1}{\sqrt{h^2}}-\frac{1}{\sqrt{R^2+h^2}}\right)\,\hat{\textbf{k}},
\end{equation}

which is the answer to the problem.

Using the numerical values in SI units (\(h=0.6\,\text{m}\) and \(Q=0.05\,\text{C}\)), we obtain

\begin{equation}
\vec{E}=\frac{(0.05\,\text{C})(0.6\,\text{m})}{2\pi(8.854\times10^{-12}\,\text{F/m})(0.2\,\text{m})^2}\left(\frac{1}{\sqrt{(0.6\,\text{m})^2}}-\frac{1}{\sqrt{(0.2\,\text{m})^2+(0.6\,\text{m})^2}}\right)\,\hat{\textbf{k}},
\end{equation}

\begin{equation}
\vec{E}\approx 1.15\times 10^{9}\,\text{N/C}\,\hat{\textbf{k}}.
\end{equation}

You need to be registered and logged in to take this quiz. Log in or Register