Solved Problem on Coulomb's Law and Electric Field
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A ring of radius a carries a uniformly distributed electric charge Q. 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;
- Charge of the ring: Q;
- Distance to the point where we want the electric field: z.
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 1-A).
\[
\begin{gather}
\mathbf{r}={\mathbf{r}}_{p}-{\mathbf{r}}_{q}
\end{gather}
\]
From the geometry of the problem, we choose cylindrical coordinates (Figure 1-B), the rq vector, 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
\[
\begin{gather}
\mathbf{r}=z\;\mathbf{k}-\left(x\;\mathbf{i}+y\;\mathbf{j}\right)\\[5pt]
\mathbf{r}=-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k} \tag{I}
\end{gather}
\]
From expression (I), the magnitude of the position vector will be
\[
\begin{gather}
r^{2}=(-x)^{2}+(-y)^{2}+z^{2}\\[5pt]
r=\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}} \tag{II}
\end{gather}
\]
where x, y, and z, in cylindrical coordinates, are given by
\[
\begin{gather}
\left\{
\begin{array}{l}
x=a\cos \theta \\
y=a\sin \theta \\
z=z
\end{array}
\right. \tag{III}
\end{gather}
\]
Solution
The electric field vector is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int{\frac{dq}{r^{2}}\;\frac{\mathbf{r}}{r}}}
\end{gather}
\]
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon _{0}}\int{\frac{dq}{r^{3}}\;\mathbf{r}} \tag{IV}
\end{gather}
\]
Using the expression of the linear density of charge λ, we have the charge element dq
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\lambda =\frac{dq}{ds}}
\end{gather}
\]
\[
\begin{gather}
dq=\lambda \;ds \tag{V}
\end{gather}
\]
where ds is an arc element with angle dθ (Figure 2)
Figure 2
\[
\begin{gather}
ds=a\;d\theta \tag{VI}
\end{gather}
\]
substituting the expression (VI) into expression (V)
\[
\begin{gather}
dq=\lambda a\;d\theta \tag{VII}
\end{gather}
\]
Substituting expressions (I), (II), and (VII) into expression (IV)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int {\frac{\lambda a\;d\theta}{\left[\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}\right]^{3}}}\left(-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\lambda a\;d\theta}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}}}\left(-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k}\right) \tag{VIII}
\end{gather}
\]
substituting expressions (III) into expression (VIII)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int {\frac{\lambda a\;d\theta}{\left[\left(a\cos \theta\right)^{2}+\left(\;a\sin \theta\right)^{2}+z^{2}\right]^{\frac{3}{2}}}}\left(-a\cos \theta\;\mathbf{i}-a\sin \theta\;\mathbf{j}+z\mathbf{k}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}} \int {\frac{\lambda a\;d\theta}{\left[a^{2}\cos^{2}\theta +a^{2}\sin ^{2}\theta+z^{2}\right]^{\frac{3}{2}}}}\left(-a\cos \theta\;\mathbf{i}-a\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\lambda a\;d\theta}{\left[a^{2}\underbrace{\left(\cos ^{2}\theta +\sin ^{2}\theta\right)}_{1}+z^{2}\right]^{\frac{3}{2}}}}\left(-a\cos \theta\;\mathbf{i}-a\sin \theta\;\mathbf{j}+z\;\mathbf{k}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\lambda a\;d\theta}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}}\left(-a\cos \theta\;\mathbf{i}-a\sin \theta\;\mathbf{j}+z\;\mathbf{k}\;\right)
\end{gather}
\]
As the charge density λ 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, we can write
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{\lambda\;a}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\left(-a\int \cos \theta\;d\theta \;\mathbf{i}-a\int \sin \theta \;d\theta\;\mathbf{j}+z\int \;d\theta\;\mathbf{k}\;\right)
\end{gather}
\]
The limits of integration will be 0 and 2π (a complete lap in the ring)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{\lambda a}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\left(-a\underbrace{\int_{0}^{2\pi}\cos \theta \;d\theta}_{0}\;\mathbf{i}-a\underbrace{\int_{0}^{2\pi}\sin \theta \;d\theta}_{0}\;\mathbf{j}+z\int_{0}^{2\pi}\;d\theta \;\mathbf{k}\right)
\end{gather}
\]
Integration of \( \displaystyle \int_{0}^{{2\pi}}\cos \theta \;d\theta \)
1st method
Figure 3
1st method
\[
\begin{gather}
\int_{0}^{{2\pi}}\cos \theta \;d\theta =\left.\sin \theta\;\right|_{\;0}^{\;2\pi}=\sin 2\pi-\sin 0=0-0=0
\end{gather}
\]
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 3).
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 3).
Integration of \( \displaystyle \int_{0}^{{2\pi}}\sin \theta \;d\theta \)
1st method
Figure 4
Figure 5
1st method
\[
\begin{gather}
\int_{0}^{{2\pi}}\sin \theta \;d\theta =\left.-\cos \theta\;\right|_{\;0}^{\;2\pi }=-(\cos 2\pi -\cos 0)=-(1-1)=0
\end{gather}
\]
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 (Figure 4).
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 (Figure 4).
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 5).
Integration of \( \displaystyle \int_{0}^{{2\pi}}d\theta \)
\[
\begin{gather}
\int_{0}^{{2\pi}}d\theta =\left.\theta \;\right|_{\;0}^{\;2\pi }=2\pi-0=2\pi
\end{gather}
\]
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{\lambda a}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}2\pi z\;\mathbf{k}\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon _{0}}\frac{2\pi \lambda a z}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\;\mathbf{k} \tag{IX}
\end{gather}
\]
The total charge of the ring is Q, and its length is 2πa, the linear charge density can
be written
\[
\begin{gather}
\lambda =\frac{Q}{2\pi a}\\[5pt]
Q=2\pi a\lambda \tag{X}
\end{gather}
\]
substituting expression (X) into expression (IX)
\[ \bbox[#FFCCCC,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{Qz}{\left(a^{2}+z^{2}\right)^{\frac{3}{2}}}\;\mathbf{k}}
\]
Figure 6
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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 .