An arc of a circle of radius a and central angle θ0 carries an electric charge
Q uniformly distributed along the arc. Determine:
a) The electric field vector, at the points of the line passing through the center of the arc and is
perpendicular to the plane containing the arc;
b) The electric field vector in the center of curvature of the arc;
c) The electric field vector when the central angle tends to zero.
Problem data:
- Radius of the arc: a;
- Central angle of the arc: θ0;
- Electric charge on the arc: Q.
Problem diagram:
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
yz plane, is written as
\( {\mathbf{r}}_{q}=y\;\mathbf{j}+z\;\mathbf{k} \)
and the
rp vector only has a component in the
i direction,
\( {\mathbf{r}}_{p}=x\;\mathbf{i} \)
(contrary to what is usually where the
rq vector is on the
xy plane and the
axis of the cylinder in the
k direction), the position vector will be
\[
\begin{gather}
\mathbf{r}=x\;\mathbf{i}-\left(y\;\mathbf{j}+z\;\mathbf{k}\right)\\[5pt]
\mathbf{r}=x\;\mathbf{i}-y\;\mathbf{j}-z\;\mathbf{k} \tag{I}
\end{gather}
\]
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
\[
\begin{gather}
\left\{
\begin{array}{l}
x=x \\
y=a\cos \theta \\
z=a\sin \theta
\end{array}
\right. \tag{III}
\end{gather}
\]
Solution
a) 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)
\[
\begin{gather}
ds=a\;d\theta \tag{VI}
\end{gather}
\]
substituting the expression (VI) into expression (V)
\[
\begin{gather}
dq=\lambda a\;d\theta \tag{VII}
\end{gather}
\]
Figure 2
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)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\lambda a\;d\theta }{\left[\;x^{2}+\left(a\cos \theta\right)^{2}+\left(a\sin \theta\right)^{2}\right]^{\;\frac{3}{2}}}}\left(x\;\mathbf{i}-a\cos\theta\;\mathbf{j}-a\sin \theta\;\mathbf{k}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\lambda a\;d\theta}{\left[x^{2}+a^{2}\cos^{2}\theta +a^{2}\sin ^{2}\theta\;\right]^{\;\frac{3}{2}} }}\left(x\;\mathbf{i}-a\cos \theta\;\mathbf{j}-a\sin \theta\;\mathbf{k}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\lambda a\;d\theta}{\left[x^{2}+a^{2}\underbrace{\left(\cos ^{2}\theta+\sin ^{2}\theta\right)}_{1}\right]^{\;\frac{3}{2}}}\left(x\;\mathbf{i}-a\cos\theta\;\mathbf{j}-a\sin \theta\;\mathbf{k}\right)}\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\lambda a\;d\theta}{\left(x^{2}+a^{2}\right)^{\;\frac{3}{2}}}\left(x\;\mathbf{i}-a\cos\theta\;\mathbf{j}-a\sin \theta\;\mathbf{k}\right)}
\end{gather}
\]
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
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{\lambda a}{\left(x^{2}+a^{2}\right)^{\;\frac{3}{2}}}\left(x\int d\theta\;\mathbf{i}-a\int \cos \theta d\theta\;\mathbf{j}-a\int \sin \theta \;d\theta\;\mathbf{k}\right)
\end{gather}
\]
As there is symmetry, we can divide the central angle θ
0 into two parts measuring
\( \frac{\theta_{0}}{2} \)
clockwise and
\( -{\frac{\theta_{0}}{2}} \)
counterclockwise (Figure 3), the integration limits will be
\( -{\frac{\theta_{0}}{2}} \)
and
\( \frac{\theta_{0}}{2} \)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{\lambda a}{\left(x^{2}+a^{2}\right)^{\;\frac{3}{2}}}\left(x\int_{-{\frac{\theta_{0}}{2}}}^{{\frac{\theta_{0}}{2}}}d\theta\;\mathbf{i}-a\int _{-{\frac{\theta_{0}}{2}}}^{{\frac{\theta_{0}}{2}}}\cos \theta \;d\theta\;\mathbf{j}-a\int_{-{\frac{\theta_{0}}{2}}}^{{\frac{\theta_{0}}{2}}}\sin \theta \;d\theta\;\mathbf{k}\right)
\end{gather}
\]
Figure 3
Integration of
\( \displaystyle \int_{-{\frac{\theta_{0}}{2}}}^{{\frac{\theta_{0}}{2}}}\;d\theta \)
\[
\begin{gather}
\int_{-{\frac{\theta_{0}}{2}}}^{{\frac{\theta_{0}}{2}}}\;d\theta=\frac{\theta_{0}}{2}-\left(-{\frac{\theta_{0}}{2}}\right)=\frac{\theta_{0}}{2}+\frac{\theta_{0}}{2}=2\frac{\theta_{0}}{2}=\theta_{0}
\end{gather}
\]
Integration of
\( \displaystyle \int_{-{\frac{\theta _{\;0}}{2}}}^{{\frac{\theta _{\;0}}{2}}}\cos\theta \;d\theta \)
\[
\begin{gather}
\int_{-{\frac{\theta_{0}}{2}}}^{{\frac{\theta_{0}}{2}}}\cos \theta\;d\theta =\left.\sin \theta \right|_{\;-\frac{\theta_{0}}{2}}^{\;\frac{\theta_{0}}{2}}=\sin \frac{\theta_{0}}{2}-\sin \;\left(-{\frac{\theta_{0}}{2}}\right)
\end{gather}
\]
the function sine is an odd function
\( f(-x)=-f(x) \),
\( \sin \left(-{\dfrac{\theta_{0}}{2}}\right)=-\sin \dfrac{\theta_{0}}{2} \)
\[
\begin{gather}
\int_{-{\frac{\theta_{0}}{2}}}^{{\frac{\theta_{0}}{2}}}\cos\theta \;d\theta =\sin \frac{\theta_{0}}{2}-\left[-\sin \;\frac{\theta_{\;0}}{2}\right]=\sin \frac{\theta_{\;0}}{2}+\sin \frac{\theta_{0}}{2}=2\;\sin \;\frac{\theta_{0}}{2}
\end{gather}
\]
Integration of
\( \displaystyle \int_{-{\frac{\theta_{;0}}{2}}}^{{\frac{\theta_{\;0}}{2}}}\sin \theta \;d\theta \)
\[
\begin{gather}
\int_{-{\frac{\theta_{0}}{2}}}^{{\frac{\theta_{0}}{2}}}\sin \theta \;d\theta =\left.-\cos \theta\right|_{-\frac{\theta_{0}}{2}}^{\;\frac{\theta_{0}}{2}}=-\left[\cos \frac{\theta_{0}}{2}-\cos \left(-{\frac{\theta_{0}}{2}}\right)\right]
\end{gather}
\]
the function cosine is an even function
\( f(x)=f(-x) \),
\( \cos \left(-{\dfrac{\theta_{0}}{2}}\right)=\cos \dfrac{\theta_{0}}{2} \)
\[
\begin{gather}
\int_{-{\frac{\theta_{0}}{2}}}^{{\frac{\theta_{0}}{2}}}\sin \theta \;d\theta =-\left[\cos \frac{\theta_{0}}{2}-\cos \frac{\theta_{0}}{2}\right]=0
\end{gather}
\]
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{\lambda a}{\left(x^{2}+a^{2}\right)^{\;\frac{3}{2}}}\left(x\theta_{0}\;\mathbf{i}-2a\sin \frac{\theta_{0}}{2}\;\mathbf{j}-0\;\mathbf{k}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\frac{\lambda a}{\left(x^{2}+a^{2}\right)^{\;\frac{3}{2}}}\left(x\theta_{0}\;\mathbf{i}-2a\sin \frac{\theta_{0}}{2}\;\mathbf{j}\right) \tag{IX}
\end{gather}
\]
Note: The integral in the
k direction is equal to zero, because an element of charge
dq, produces at a point, an element of the field that can be decomposed in the elements,
dEx, −
dEy and
−
dEz (Figure 4-A). Another element of charge placed in a symmetrical
position produces at the same point another element of the field that can be decomposed in the elements
dEx, −
dEy and
dEz (Figure 4-B), so the elements in the
k direction cancel and
only elements in the
i and
j directions contribute to the total field.
The total charge of the arc is
Q, and its length is
aθ
0, so the linear
density of charge can be written
\[
\begin{gather}
\lambda =\frac{Q}{a\theta_{0}} \tag{X}
\end{gather}
\]
substituting expression (X) into expression (IX)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{Q}{a\theta_{0}}\frac{a}{\left(x^{2}+a^{2}\right)^{\;\frac{3}{2}}}\left(x\theta_{0}\;\mathbf{i}-2a\sin \frac{\theta_{0}}{2}\;\mathbf{j}\right)\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\left(x^{2}+a^{2}\right)^{\;\frac{3}{2}}}\left(\frac{1}{\theta_{0}}x\theta_{0}\;\mathbf{i}-\frac{1}{\theta_{0}}2a\sin \frac{\theta_{0}}{2}\;\mathbf{j}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{Q}{\left(x^{2}+a^{2}\right)^{\;\frac{3}{2}}}\left(x\;\mathbf{i}-\frac{2a}{\theta_{0}}\sin \frac{\theta_{0}}{2}\;\mathbf{j}\right)}
\end{gather}
\]
Figure 5
b) In the center of curvature, we have
x = 0, substituting in the solution of the previous item
(Figure 6)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{Q}{\left(0^{2}+a^{2}\right)^{\;\frac{3}{2}}}\left(0\;\mathbf{i}-\frac{2a}{\theta_{0}}\sin \frac{\theta_{0}}{2}\;\mathbf{j}\right)\\[5pt]
\mathbf{E}=\frac{-{1}}{4\pi\epsilon _{0}}\frac{Q}{a^{3}}\frac{2a}{\theta_{0}}\sin \frac{\theta_{0}}{2}\;\mathbf{j}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\mathbf{E}=\frac{-{1}}{4\pi \epsilon_{0}}\frac{Q}{a^{2}}\frac{2}{\theta_{0}}\sin \frac{\theta_{0}}{2}\;\mathbf{j}}
\end{gather}
\]
Figure 6
c) When the central angle tends to zero
(
\( \theta_{0}\rightarrow 0 \)),
the arc tends to a point charge, applying the limit to the solution of the previous item (Figure 7)
\[
\begin{gather}
\mathbf{E}=\underset{\theta_{0}\rightarrow0}{\lim }-\frac{1}{4\pi \epsilon_{0}}\;\frac{Q}{a^{2}}\frac{2}{\theta_{0}}\sin \frac{\theta_{\;0}}{2}\;\mathbf{j}
\end{gather}
\]
turning the term
\( \frac{2}{\theta_{0}} \)
upside down
\[
\begin{gather}
\mathbf{E}=\underset{\theta_{0}\rightarrow 0}{\lim}-{\frac{1}{4\pi \epsilon_{0}}\frac{Q}{a^{2}}\dfrac{\sin \dfrac{\theta_{0}}{2}}{\dfrac{\theta_{\;0}}{2}}\;\mathbf{j}}
\end{gather}
\]
Figure 7
From the limit
\( \underset{x\rightarrow 0}{\lim }{\dfrac{\sin x}{x}}=1 \)
\[
\begin{gather}
\mathbf{E}=-{\frac{1}{4\pi \epsilon_{0}}}\frac{Q}{a^{2}}\underbrace{\;\underset{\theta_{0}\rightarrow 0}{\lim }{\dfrac{\sin \dfrac{\theta_{0}}{2}}{\dfrac{\theta_{0}}{2}}}}_{1}\;\mathbf{j}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\mathbf{E}=-{\frac{Q}{4\pi \epsilon_{0}a^{2}}}\;\mathbf{j}}
\end{gather}
\]
and the result is reduced to the electric field vector of a point charge.