Solved Problem on Coulomb's Law and Electric Field
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A ring of radius a carries a charge whose density varies directly proportional to the angular
position between the points zero and 2π there is a very thin membrane of an insulator material
separating these two points. Calculate the electric field vector at a point P on the
symmetry axis perpendicular to the plane of the ring at a distance z from its center.
Problem data:
- Radius of the ring: a;
- Distance to the point where we want the electric field: z.
The linear density of charge of the ring is directly proportional to the angular position of the charge
(Figure 1)
\[
\begin{gather}
\lambda (\theta )=\alpha \theta \tag{I}
\end{gather}
\]
where α is a constant that makes expression dimensionally consistent. The points zero and 2π
represent the same point, this means that there are two different densities of charge to the same point.
To solve this inconsistency the problem tells us, that there is a very thin insulator membrane, so
physically the two points with different charges are separated and mathematically we can integrate from
zero to 2π.
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).
\[
\begin{gather}
\mathbf{r}={\mathbf{r}}_{p}-{\mathbf{r}}_{q}
\end{gather}
\]
From the geometry of the problem, we choose cylindrical coordinates (Figure 2-B), the rq vector, which is on the xy plane, is written as \( {\mathbf{r}}_{q}=x\;\mathbf{i}+y\;\mathbf{j} \) and the vector rp only has a component in the k direction, and the vector rp only has a component in the k direction k, \( {\mathbf{r}}_{p}=z\;\mathbf{k} \), the position vector will be
\[
\begin{gather}
\mathbf{r}=z\;\mathbf{k}-\left(x\;\mathbf{i}+y\;\mathbf{j}\right)\\[5pt]
\mathbf{r}=-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k} \tag{II}
\end{gather}
\]
From expression (I), the magnitude of the position vector r will be
\[
\begin{gather}
r^{2}=(-x)^{2}+(-y)^{2}+z^{2}\\[5pt]
r=\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}} \tag{III}
\end{gather}
\]
where x, y, and z, in cylindrical coordinates, are given by
\[
\begin{gather}
\left\{
\begin{array}{l}
x=a\cos \theta \\
y=a\sin \theta \\
z=z
\end{array}
\right. \tag{IV}
\end{gather}
\]
Solution
The electric field vector is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int{\frac{dq}{r^{2}}\;\frac{\mathbf{r}}{r}}}
\end{gather}
\]
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int{\frac{dq}{r^{3}}\;\mathbf{r}} \tag{V}
\end{gather}
\]
Using the expression of the linear density of charge λ, we have the charge element dq
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\lambda (\theta )=\frac{dq}{ds}}
\end{gather}
\]
\[
\begin{gather}
dq=\lambda (\theta )\;ds \tag{VI}
\end{gather}
\]
Figure 3
where ds is an arc element with angle dθ (Figure 3)
\[
\begin{gather}
ds=a\;d\theta \tag{VII}
\end{gather}
\]
substituting expressions (I) and (VII) into expression (VI)
\[
\begin{gather}
dq=\alpha \theta a\;d\theta \tag{VIII}
\end{gather}
\]
Substituting expressions (I), (III), and (VIII) into expression (V)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int {\frac{\alpha \theta a\;d\theta}{\left[\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}\right]^{3}}}\left(-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\alpha \theta a\;d\theta}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}}}\left(-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k}\right) \tag{IX}
\end{gather}
\]
substituting expressions (IV) into expression (IX)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int {\frac{\alpha \theta a\;d\theta }{\left[\left(a\cos \theta\right)^{2}+\left(a\sin \theta\right)^{2}+z^{2}\right]^{\frac{3}{2}}}\left(-a\cos \theta\;\mathbf{i}-a\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)}\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\alpha \theta a\;d\theta }{\left[a^{2}\cos^{2}\theta +a^{2}\sin ^{2}\theta+z^{2}\right]^{\frac{3}{2}}}\left(-a\cos \theta\;\mathbf{i}-a\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)}\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\alpha \theta a\;d\theta}{\left[a^{2}\underbrace{\left(\cos ^{2}\theta+\sin ^{2}\theta\right)}_{1}+z^{2}\;\right]^{\frac{3}{2}}}\left(-a\cos \theta\;\mathbf{i}-a\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)}\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\alpha \theta a\;d\theta}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\left(-a\cos \theta\;\mathbf{i}-a\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)}
\end{gather}
\]
As the constant of proportionality α and the radius a are constants they are moved outside of
the integral, and the integral of the sum is equal to the sum of the integrals
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{\alpha a}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\left(-a\int \theta \cos\theta \;d\theta \;\mathbf{i}-a\int \theta\sin \theta \;d\theta \;\mathbf{j}+z\int\theta \;d\theta \;\mathbf{k}\right)
\end{gather}
\]
The limits of integration will be 0 and 2π (a complete lap in the ring)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{\alpha a}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\left(-a\int _{0}^{{2\pi}}\theta \cos \theta \;d\theta \;\mathbf{i}-a\int_{0}^{{2\pi }}\theta \sin \theta \;d\theta\;\mathbf{j}+z\int _{0}^{{2\pi }}\theta \;d\theta\;\mathbf{k}\right)
\end{gather}
\]
Integration of \( \displaystyle \int _{0}^{{2\pi }}\theta \cos \theta \;d\theta \)
Using Integration by Parts \( \int uv'=uv-\int u'v \), we chose
Using Integration by Parts \( \int uv'=uv-\int u'v \), we chose
\[
\begin{align}
u=\theta \qquad \qquad & v'=\cos \theta\\[5pt]
u'=1\qquad \qquad & v=\sin \theta
\end{align}
\]
\[
\begin{gather}
\int _{0}^{{2\pi }}\theta \cos \theta \;d\theta=\theta \sin \theta \;|_{\;0}^{\;2\pi }-\int _{0}^{{2\pi}}\sin \theta \;d\theta \\[5pt]
\int _{0}^{{2\pi }}\theta \cos\theta \;d\theta =\theta \sin \theta \;|_{\;0}^{\;2\pi}-\left(-\cos \theta \;|_{\;0}^{\;2\pi }\right)\\[5pt]
\int _{0}^{{2\pi}}\theta \cos \theta \;d\theta =\theta \sin \theta\;|_{\;0}^{\;2\pi }+\cos \theta \;|_{\;0}^{\;2\pi }\\[5pt]
\int _{0}^{{2\pi}}\theta \cos \theta \;d\theta =\left(2\pi .\sin 2\pi-0.\sin 0\right)+\left(\cos 2\pi -\cos 0\right)\\[5pt]
\int_{0}^{{2\pi }}\theta \cos \theta \;d\theta =\left(2\pi.0-0.0\right)+\left(1-1\right)\\[5pt]
\int _{0}^{{2\pi }}\theta \cos \theta\;d\theta =0
\end{gather}
\]
Integration of \( \displaystyle \int _{0}^{{2\pi }}\theta \sin \theta \;d\theta \)
Using Integration by Parts \( \int uv'=uv-\int u'v \), we chose
Using Integration by Parts \( \int uv'=uv-\int u'v \), we chose
\[
\begin{align}
u=\theta\qquad \qquad & v\acute{}=\sin \theta\\[5pt]
u\acute{}=1\qquad & \qquad v=-\cos \theta
\end{align}
\]
\[
\begin{gather}
\int _{0}^{{2\pi }}\theta \sin \theta\;d\theta =-\theta \cos \theta \;|_{\;0}^{\;2\pi }-\int _{0}^{{2\pi}}-\cos \theta \;d\theta \\[5pt]
\int _{0}^{{2\pi }}\theta\sin \theta \;d\theta =-\theta \cos \theta\;|_{\;0}^{\;2\pi }+\int _{0}^{{2\pi }}\cos \theta \;d\theta \\[5pt]
\int_{0}^{{2\pi }}\theta \sin \theta \;d\theta =-\theta \cos\theta \;|_{\;0}^{\;2\pi }+\sin \theta \;|_{\;0}^{\;2\pi}\\[5pt]
\int _{0}^{{2\pi }}\theta \sin \theta \;d\theta=-\left(2\pi .\cos 2\pi -0.\cos 0\right)+\left(\sin 2\pi-\sin 0\right)\\[5pt]
\int _{0}^{{2\pi }}\theta\sin \theta \;d\theta =-\left(2\pi.1-0.1\right)+\left(0-0\right)\\[5pt]
\int _{0}^{{2\pi }}\theta\sin \theta \;d\theta =-2\pi
\end{gather}
\]
Integration of \( \displaystyle \int _{0}^{{2\pi }}\theta \;d\theta \)
\[
\begin{gather}
\int_{0}^{{2\pi}}\theta \;d\theta =\left.\frac{\theta^{2}}{2}\;\right|_{\;0}^{\;2\pi }=\frac{(2\pi)^{2}}{2}-\frac{0^{2}}{2}=\frac{4\pi^{2}}{2}-0=2\pi^{2}
\end{gather}
\]
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{\alpha a}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\left[-a.0\;\mathbf{i}-a(-2\pi)\;\mathbf{j}+z\;(2\pi^{2})\;\mathbf{k}\right]\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\frac{\alpha a}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\left(2\pi a\;\mathbf{j}+2\pi^{2}z\;\mathbf{k}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\frac{2\pi \alpha a}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\left(a\;\mathbf{j}+\pi z\;\mathbf{k}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\mathbf{E}=\frac{1}{2\epsilon_{0}}\frac{\alpha a}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\left(a\;\mathbf{j}+\pi z\;\mathbf{k}\right)}
\end{gather}
\]
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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 .