Solved Problem on Coulomb's Law and Electric Field
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A disk of radius a carries a charge whose density varies directly proportional to the radial
position. Calculate the electric field vector at a point P on the symmetry axis perpendicular to
the plane of the disk at a distance z of its center.
Problem data:
- Radius of disk: a;
- Distance to the point where we want the electric field: z.
The surface density of charges is directly proportional to the radial position of charges (Figure 1)
\[
\begin{gather}
\sigma (r)=\alpha r \tag{I}
\end{gather}
\]
where α is a constant that makes expression dimensionally consistent. So in the center of the
disk where the radius is zero the density of charge is zero, at the edge where the radius is equal to
a the density of charge is αa.
Figure 1
The position vector r goes from an element of charge dq to point P where we want to calculate the electric field, the vector rq locates the charge element relative to the origin of the reference frame, and the vector rp locates point P (Figure 2-A).
\[
\mathbf{r}={\mathbf{r}}_{p}-{\mathbf{r}}_{q}
\]
The vector rq, which is on the xy plane, is written as \( {\mathbf{r}}_{q}=x\;\mathbf{i}+y\;\mathbf{j} \) and the rp vector only has a component in the k direction, \( {\mathbf{r}}_{p}=z\;\mathbf{k} \), the position vector will be (Figure 2-B)
\[
\begin{gather}
\mathbf{r}=z\;\mathbf{k}-\left(x\;\mathbf{i}+y\;\mathbf{j}\right)\\
\mathbf{r}=-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k} \tag{II}
\end{gather}
\]
From expression (II), the magnitude of the position vector r will be
\[
\begin{gather}
r^{2}=x^{2}+y^{2}+z^{2}\\
r=\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}} \tag{III}
\end{gather}
\]
setting x, y, and z (Figure 2-B)
\[
\left\{
\begin{array}{l}
x=r_{q}\cos \theta \\
y=r_{q}\sin \theta \\
z=z
\end{array}
\right. \tag{IV}
\]
Solution
The electric field vector is given by
\[ \bbox[#99CCFF,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int{\frac{dq}{r^{2}}\;\frac{\mathbf{r}}{r}}}
\]
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int{\frac{dq}{r^{3}}\;\mathbf{r}} \tag{V}
\end{gather}
\]
Using the expression of the surface density of charge σ, we have the charge element dq
\[ \bbox[#99CCFF,10px]
{\sigma (r)=\frac{dq}{dA}}
\]
\[
\begin{gather}
dq=\sigma (r)\;dA \tag{VI}
\end{gather}
\]
where dA is an element of area with angle dθ of the disk (Figure 3)
\[
\begin{gather}
dA=r_{q}\;dr_{q}\;d\theta \tag{VII}
\end{gather}
\]
Figure 3
substituting expressions (I) and (VII) into expression (VI)
\[
\begin{gather}
dq=\alpha r_{q}^{2}\;dr_{q}\;d\theta \tag{VIII}
\end{gather}
\]
Substituting expressions (II), (III), and (VII) into expression (V), and as the integration is made on the
surface of the disk, it depends on two variables r and θ, we have a double integral
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\alpha r_{q}^{2}\;dr_{q}\;d\theta}{\left[\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}\right]^{3}}}\left(-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k}\right)\\
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\alpha r_{q}^{2}\;dr_{q}\;d\theta}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}}}\left(-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k}\right) \tag{IX}
\end{gather}
\]
substituting expressions (IV) into expression (VIII)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\iint {\frac{\alpha r_{q}^{2}\;dr_{q}\;d\theta}{\left[\left(r_{q}\cos \theta\right)^{2}+\left(r_{q}\sin \theta\right)^{2}+z^{2}\right]^{\frac{3}{2}}}\left(-r_{q}\cos \theta\;\mathbf{i}-r_{q}\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)}\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\alpha r_{q}^{2}\;dr_{q}\;d\theta}{\left[r_{q}^{2}\cos ^{2}\theta +r_{q}^{2}\sin ^{2}\theta+z^{2}\right]^{\frac{3}{2}}}\left(-r_{q}\cos \theta\;\mathbf{i}-r_{q}\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)}\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\alpha r_{q}^{2}\;dr_{q}\;d\theta}{\left[r_{q}^{2}\left(\underbrace{\cos ^{2}\theta+\sin ^{2}\theta}_{1}\right)+z^{2}\right]^{\frac{3}{2}}}\left(-r_{q}\cos \theta\;\mathbf{i}-r_{q}\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)}\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\alpha r_{q}^{2}\;dr_{q}\;d\theta}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}\left(-r_{q}\cos \theta\;\mathbf{i}-r_{q}\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)}
\end{gather}
\]
The constant of proportionality σ is moved outside of the integral, and the integral
of the sum is equal to the sum of the integrals
\[
\mathbf{E}=\frac{\sigma}{4\pi \epsilon_{0}}\left(-\iint {\frac{r_{q}^{3}\cos \theta \;dr_{q}\;d\theta}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\;\mathbf{i}-\iint {\frac{r_{q}^{3}\;\sin \theta \;dr_{q}\;d\theta}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\;\mathbf{j}+\iint {\frac{z r_{q}\;dr_{q}\;d\theta}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\mathbf{k}\right)
\]
The limits of integration will be 0 and a in drq, along the disk radius, 0 and
2π in dθ, a complete lap in the disk, the integral can be separated
\[
\begin{split}
\mathbf{E}=& \frac{\alpha}{4\pi \epsilon_{0}}\left(-\int_{0}^{a}{\frac{r_{q}^{3}\;dr_{q}}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\underbrace{\int_{0}^{2\pi}{\cos \theta \;d\theta}}_{0}\;\mathbf{i}-\right.\\
& \left. -\int_{0}^{a}{\frac{r_{q}^{3}\;dr}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\underbrace{\int_{0}^{2\pi}{\sin \theta \;d\theta}}_{0}\;\mathbf{j}+z\int_{0}^{a}{\frac{r_{q}\;dr_{q} }{\left(r^{2}+z^{2}\right)^{\frac{3}{2}}}}\int_{0}^{2\pi}{d\theta}\;\mathbf{k}\right)
\end{split}
\]
Integration of \( \displaystyle \int_{0}^{2\pi}\cos \theta \;d\theta \)
1st method
Figure 4
1st method
\[
\begin{split}
\int_{0}^{2\pi}\cos \theta \;d\theta &=\left.\sin \theta\;\right|_{\;0}^{\;2\pi }=\sin 2\pi-\sin 0=\\
&=0-0=0
\end{split}
\]
2nd method
The graph of cosine between 0 and 2π, has a "positive" area above the x-axis between 0 and \( \frac{\pi}{2} \) and between \( \frac{3\pi}{2} \) and 2π, and a "negative" area below the x-axis between \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \), these two areas cancel in the integration and the integral is equal to zero (Figure 4).
The graph of cosine between 0 and 2π, has a "positive" area above the x-axis between 0 and \( \frac{\pi}{2} \) and between \( \frac{3\pi}{2} \) and 2π, and a "negative" area below the x-axis between \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \), these two areas cancel in the integration and the integral is equal to zero (Figure 4).
Integration of \( \displaystyle \int_{0}^{2\pi}\sin \theta \;d\theta \)
1st method
Figure 5
Figure 6
1st method
\[
\begin{split}
\int_{0}^{2\pi}\sin \theta \;d\theta &=\left.-\cos \theta\;\right|_{\;0}^{\;2\pi }=-(\cos 2\pi -\cos 0)=\\
&=-(1-1)=0
\end{split}
\]
2nd method
The graph of sine between 0 and 2π, has a "positive" area above the x-axis between 0 and π and a "negative" area below the x-axis between π and 2π, these two areas cancel in the integration and the integral is equal to zero in i and j directions (Figure 5).
The graph of sine between 0 and 2π, has a "positive" area above the x-axis between 0 and π and a "negative" area below the x-axis between π and 2π, these two areas cancel in the integration and the integral is equal to zero in i and j directions (Figure 5).
Note: The two integrals, in directions i and j which are zero, represent
the mathematical calculation for the assertion that is usually done that the components of the
electric field parallel to the xy plane, dEP, cancel. Only
normal components to the plane dEN contribute to the total electric
field (Figure 6). As the integrals in sine and cosine are zero, we do not have to calculate the
integral of the radius.
Figure 6
Integration of \( \displaystyle \int_{{0}}^{a}{\frac{r_{q}\;dr_{q}}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}} \)
Changing the variable
for r = 0
we have \( u=0^{2}+z^{2}\Rightarrow u=z^{2} \)
for r = a
we have \( u=a^{2}+z^{2} \)
Changing the variable
\[
\begin{array}{l}
u=r_{q}^{2}+z^{2}\\
\dfrac{du}{dr_{q}}=2r_{q}\Rightarrow dr_{q}=\dfrac{du}{2r_{q}}
\end{array}
\]
changing the limits of integration
for r = 0
we have \( u=0^{2}+z^{2}\Rightarrow u=z^{2} \)
for r = a
we have \( u=a^{2}+z^{2} \)
\[
\begin{split}
\int_{z^{2}}^{{a^{2}+z^{2}}}{\frac{r_{q}}{u^{\frac{3}{2}}}\;\frac{du}{2r_{q}}} &\Rightarrow\frac{1}{2}\int_{z^{2}}^{{a^{2}+z^{2}}}{\frac{1}{u^{\frac{3}{2}}}\;du}\Rightarrow\frac{1}{2}\left.\frac{u^{-{\frac{3}{2}+1}}}{-{\frac{3}{2}+1}}\;\right|_{\;z^{2}}^{\;a^{2}+z^{2}}\Rightarrow\\[5pt]
&\Rightarrow\frac{1}{2}\left.\frac{u^{\frac{-3+2}{2}}}{\frac{-{3+2}}{2}}\;\right|_{\;z^{2}}^{\;a^{2}+z^{2}}\Rightarrow\frac{1}{2}\left.\frac{u^{-{\frac{1}{2}}}}{-{\frac{1}{2}}}\;\right|_{\;z^{2}}^{\;a^{2}+z^{2}}\Rightarrow\\[5pt]
&\Rightarrow\left.-u^{-{\frac{1}{2}}}\;\right|_{\;z^{2}}^{\;a^{2}+z^{2}}\Rightarrow\left.-{\frac{1}{u^{\frac{1}{2}}}}\;\right|_{\;z^{2}}^{\;a^{2}+z^{2}}\Rightarrow\\[5pt]
&\Rightarrow-\left(\frac{1}{\sqrt{a^{2}+z^{2}}}-\frac{1}{\sqrt{z^{2}}}\right)\Rightarrow\frac{1}{z}-\frac{1}{\sqrt{a^{2}+z^{2}}}
\end{split}
\]
Integration of \( \displaystyle \int_{{0}}^{{2\pi}}\;d\theta \)
\[
\int_{{0}}^{{2\pi}}\;d\theta =\left.\theta \;\right|_{\;0}^{\;2\pi}=2\pi -0=2\pi
\]
\[
\begin{gather}
\mathbf{E}=\frac{\alpha}{4\pi \epsilon_{0}}\left[0\;\mathbf{i}-0\;\mathbf{j}+z\;\left(\frac{1}{z}-\frac{1}{\sqrt{a^{2}+z^{2}\;}}\right)2\pi\;\mathbf{k}\right]\\
\mathbf{E}=\frac{\alpha}{\cancelto{2}{4}\cancel{\pi} \epsilon_{0}}\left[\;\left(\frac{z}{z}-\frac{z}{\sqrt{a^{2}+z^{2}}}\right)\cancel{2}\cancel{\pi}\;\mathbf{k}\right]
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{{\mathbf{E}}=\frac{\alpha}{2\epsilon_{0}}\;\left(1-\frac{z}{\sqrt{\;a^{2}+z^{2}\;}}\right)\;{\mathbf{k}}}
\]
Figure 7
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .