A ring of radius a carries a charge whose density varies directly proportional to the angular
position. Between points 0 and 2π, there is a very thin membrane of an insulating material
separating these two points. Calculate the electric field vector at 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.
Problem diagram:
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 the expression dimensionally consistent. The points 0 and 2π
represent the same point, this means there are two different densities of charge at the same point.
To solve this inconsistency, the problem tells us, that there is a very thin insulating membrane, so
physically the two points with different charges are separated, and mathematically we can integrate from
0 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,
\( {\mathbf r}_p=z\;\mathbf k \),
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
\[
\begin{gather}
r^2=(-x)^2+(-y)^2+z^2 \\[5pt]
r=\left(x^2+y^2+z^2\right)^{1/2} \tag{III}
\end{gather}
\]
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 for the linear density of charge λ, we have the charge element dq.
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\lambda(\theta)=\frac{dq}{ds}}
\end{gather}
\]
\[
\begin{gather}
dq=\lambda(\theta)\;ds \tag{VI}
\end{gather}
\]
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).
\[
\begin{gather}
\mathbf E=\frac{1}{4\pi\epsilon_0}\int{\frac{\alpha\theta a\;d\theta}{\left[\left(x^2+y^2+z^2\right)^{1/2}\right]^3}}\left(-x\;\mathbf i-y\;\mathbf j+z\;\mathbf k\right) \\[5pt]
\mathbf E=\frac{1}{4\pi\epsilon_0}\int{\frac{\alpha\theta a\;d\theta}{\left(x^2+y^2+z^2\right)^{3/2}}}\left(-x\;\mathbf i-y\;\mathbf j+z\;\mathbf k\right) \tag{IX}
\end{gather}
\]
substituting expressions (IV) into expression (IX).
\[
\begin{gather}
\mathbf E=\frac{1}{4\pi\epsilon_0}\int{\frac{\alpha\theta a\;d\theta}{\left[\left(a\cos\theta\right)^2+\left(a\sin\theta\right)^2+z^2\right]^{3/2}}\left(-a\cos\theta\;\mathbf i-a\sin\theta\;\mathbf j+z\;\mathbf k\right)} \\[5pt]
\mathbf E=\frac{1}{4\pi\epsilon_0}\int{\frac{\alpha\theta a\;d\theta}{\left[a^2\cos^2\theta+a^2\sin^2\theta+z^2\right]^{3/2}}\left(-a\cos\theta\;\mathbf i-a\sin\theta\;\mathbf j+z\;\mathbf k\right)} \\[5pt]
\mathbf E=\frac{1}{4\pi\epsilon_0}\int{\frac{\alpha\theta a\;d\theta}{\left[a^2\underbrace{\left(\cos^2\theta+\sin^2\theta\right)}_{1}+z^2\;\right]^{3/2}}\left(-a\cos\theta\;\mathbf i-a\sin\theta\;\mathbf j+z\;\mathbf k\right)} \\[5pt]
\mathbf E=\frac{1}{4\pi\epsilon_0}\int{\frac{\alpha\theta a\;d\theta}{\left(a^2+z^2\right)^{3/2}}\left(-a\cos\theta\;\mathbf i-a\sin\theta\;\mathbf j+z\;\mathbf k\right)}
\end{gather}
\]
Since the constant of proportionality α and the radius a are constants, they are moved outside
the integral, and the integral of the sum is equal to the sum of the integrals.
\[
\begin{gather}
\mathbf E=\frac{1}{4\pi\epsilon_0}\frac{\alpha a}{\left(a^2+z^2\right)^{3/2}}\left(-a\int\theta\cos\theta\;d\theta\;\mathbf i-a\int\theta\sin\theta\;d\theta\;\mathbf j+z\int\theta\;d\theta\;\mathbf k\right)
\end{gather}
\]
The limits of integration will be 0 and 2π (a complete lap in the ring).
\[
\begin{gather}
\mathbf E=\frac{1}{4\pi\epsilon_0}\frac{\alpha a}{\left(a^2+z^2\right)^{3/2}}\left(-a\int_{0}^{2\pi}\theta\cos\theta\;d\theta\;\mathbf i-a\int_{0}^{2\pi}\theta\sin\theta\;d\theta\;\mathbf j+z\int_{0}^{2\pi}\theta\;d\theta\;\mathbf k\right)
\end{gather}
\]
Integration of
\( \displaystyle \int_{0}^{2\pi}\theta\cos\theta\;d\theta\)
Using
Integration by Parts
\( \int uv'=uv-\int u'v \),
we chose
\[
\begin{align}
u=\theta\qquad \qquad & v'=\cos\theta\\[5pt]
u'=1\qquad \qquad & v=\sin\theta
\end{align}
\]
\[
\begin{gather}
\int_{0}^{2\pi}\theta\cos\theta\;d\theta=\theta\sin\theta\;|_{\;0}^{\;2\pi}-\int_{0}^{2\pi}\sin\theta\;d\theta\\[5pt]
\int_{0}^{2\pi}\theta\cos\theta\;d\theta=\theta\sin\theta\;|_{\;0}^{\;2\pi}-\left(-\cos\theta\;|_{\;0}^{\;2\pi}\right) \\[5pt]
\int_{0}^{2\pi}\theta\cos\theta\;d\theta=\theta\sin\theta\;|_{\;0}^{\;2\pi}+\cos\theta\;|_{\;0}^{\;2\pi} \\[5pt]
\int_{0}^{2\pi}\theta\cos\theta\;d\theta=\left(2\pi\times\sin 2\pi-0\times\sin 0\right)+\left(\cos 2\pi -\cos 0\right) \\[5pt]
\int_{0}^{2\pi}\theta\cos\theta\;d\theta=\left(2\pi\times 0-0\times 0\right)+\left(1-1\right) \\[5pt]
\int_{0}^{2\pi}\theta\cos\theta\;d\theta=0
\end{gather}
\]
Integration of
\( \displaystyle \int_{0}^{2\pi}\theta\sin\theta\;d\theta\)
Using
Integration by Parts
\( \int uv'=uv-\int u'v \),
we chose
\[
\begin{align}
u=\theta\qquad \qquad & v\acute{}=\sin\theta\\[5pt]
u'=1\qquad \qquad & v=-\cos\theta
\end{align}
\]
\[
\begin{gather}
\int_{0}^{2\pi}\theta\sin\theta\;d\theta=-\theta\cos\theta\;|_{\;0}^{\;2\pi}-\int_{0}^{2\pi}-\cos\theta\;d\theta \\[5pt]
\int_{0}^{2\pi}\theta\sin\theta\;d\theta=-\theta\cos\theta\;|_{\;0}^{\;2\pi}+\int_{0}^{2\pi}\cos\theta\;d\theta \\[5pt]
\int_{0}^{2\pi}\theta\sin\theta\;d\theta=-\theta\cos\theta\;|_{\;0}^{\;2\pi}+\sin\theta\;|_{\;0}^{\;2\pi} \\[5pt]
\int_{0}^{2\pi}\theta\sin\theta\;d\theta=-\left(2\pi\times\cos 2\pi -0\times\cos 0\right)+\left(\sin 2\pi-\sin 0\right) \\[5pt]
\int_{0}^{2\pi}\theta\sin\theta\;d\theta=-\left(2\pi\times 1-0\times 1\right)+\left(0-0\right) \\[5pt]
\int_{0}^{2\pi}\theta\sin\theta\;d\theta=-2\pi
\end{gather}
\]
Integration of
\( \displaystyle \int_{0}^{2\pi}\theta\;d\theta \)
\[
\begin{gather}
\int_{0}^{2\pi}\theta\;d\theta=\left.\frac{\theta^2}{2}\;\right|_{\;0}^{\;2\pi}=\frac{(2\pi)^2}{2}-\frac{0^2}{2}=\frac{4\pi^2}{2}-0=2\pi^2
\end{gather}
\]
\[
\begin{gather}
\mathbf E=\frac{1}{4\pi\epsilon_0}\frac{\alpha a}{\left(a^2+z^2\right)^{3/2}}\left[-a\times 0\;\mathbf i-a(-2\pi)\;\mathbf j+z\;(2\pi^2)\;\mathbf k\right] \\[5pt]
\mathbf E=\frac{1}{4\pi\epsilon_0}\frac{\alpha a}{\left(a^2+z^2\right)^{3/2}}\left(2\pi a\;\mathbf j+2\pi^2z\;\mathbf k\right) \\[5pt]
\mathbf E=\frac{1}{4\pi\epsilon_0}\frac{2\pi\alpha a}{\left(a^2+z^2\right)^{3/2}}\left(a\;\mathbf j+\pi z\;\mathbf k\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\mathbf E=\frac{1}{2\epsilon_0}\frac{\alpha a}{\left(a^2+z^2\right)^{3/2}}\left(a\;\mathbf j+\pi z\;\mathbf k\right)}
\end{gather}
\]
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