Solved Problem on Coulomb's Law and Electric Field

A disk of radius a has at the center a hole of radius b and carries a uniformly distributed charge Q. Calculate the electric field vector at a point P on the symmetry axis perpendicular to the plane of the disk at a distance z from its center.

Problem data:

Radius of the disk: a;
Radius of hole: b;
Charge of the disk: Q;
Distance to the point where we want the electric field: z.

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 r_q locates the charge element relative to the origin of the reference frame, and the vector r_p locates point P (Figure 1-A).

\[ \begin{gather} \mathbf{r}={\mathbf{r}}_{p}-{\mathbf{r}}_{q} \end{gather} \]

Figure 1

Based on the geometry of the problem, we choose cylindrical coordinates (Figure 1-B), the vector r_q only has a component in the e_r direction, \( {\mathbf{r}}_{q}=r_{q}\;\mathbf{e}_{r} \) and the vector r_p only has a component in the z direction, \( {\mathbf{r}}_{p}=r_{p}\;\mathbf{e}_{z} \). Converting cylindrical coordinates to Cartesian coordinates x, y and z are given by

\[ \begin{gather} \left\{ \begin{array}{l} x=r_{q}\cos \theta \\ y=r_{q}\operatorname{sen}\theta \\ z=z \end{array} \right. \tag{I} \end{gather} \]

Note: In the Figure 1-B, i, j e k are unit vectors in the Cartesian coordinates, and e_r, e_θ e e_z are unit vectors in the cylindrical coordinates.

After the conversion the vector r_q, is written as \( {\mathbf{r}}_{q}=x\;\mathbf{i}+y\;\mathbf{j} \), and the vector r_p as \( {\mathbf{r}}_{p}=z\;\mathbf{k} \).
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 (II), the magnitude of the position vector r will be

\[ \begin{gather} r^{2}=(-x)^{2}+(-y)^{2}+z^{2}\\[5pt] r=\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}} \tag{III} \end{gather} \]

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 surface density of charge σ, we have the charge element dq

\[ \begin{gather} \bbox[#99CCFF,10px] {\sigma =\frac{dq}{dA}} \end{gather} \]

\[ \begin{gather} dq=\sigma \;dA \tag{V} \end{gather} \]

where dA is an element of area with angle dθ of the disk (Figure 2)

\[ \begin{gather} dA=r_{q}\;dr_{q}\;d\theta \tag{VI} \end{gather} \]

Figure 2

substituting the expression (VI) into expression (V)

\[ \begin{gather} dq=\sigma r_{q}\;dr_{q}\;d\theta \tag{VII} \end{gather} \]

Substituting expressions (II), (II), and (VII) into expression (IV), and as the integration is made on the surface of the disk, it depends on two variables r_q and θ, we have a double integral

\[ \begin{gather} \mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\iint {\frac{\sigma r_{q}\;dr_{q}\;d\theta}{\left[\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}\right]^{3}}}\left(-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k}\right)\\[5pt] \mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\sigma r_{q}\;dr_{q}\;d\theta}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}}}\left(-x\;\mathbf{i}-y\;\mathbf{j}+z\;\mathbf{k}\right) \tag{VIII} \end{gather} \]

substituting expressions (I) into expression (VIII)

\[ \begin{gather} \mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\sigma r_{q}\;dr_{q}\;d\theta}{\left[\left(r_{q}\cos \theta\right)^{2}+\left(r_{q}\sin\theta\right)^{2}+z^{2}\right]^{\frac{3}{2}}}\left(-r_{q}\cos \theta\;\mathbf{i}-r_{q}\sin\theta\;\mathbf{j}+z\mathbf{k}\right)}\\[5pt] \mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\sigma r_{q}\;dr_{q}\;d\theta}{\left[r_{q}^{2}\cos^{\;2}\theta+r_{q}^{2}\sin^{2}\theta+z^{2}\;\right]^{\frac{3}{2}}}\left(-r_{q}\cos \theta\;\mathbf{i}-r_{q}\sin\theta\;\mathbf{j}+z\;\mathbf{k}\right)}\\[5pt] \mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\sigma r_{q}\;dr_{q}\;d\theta}{\left[r_{q}^{2}\underbrace{\left(\cos^{2}\theta+\sin^{2}\theta\right)}_{1}+z^{2}\right]^{\frac{3}{2}}}\left(-r_{q}\cos \theta\;\mathbf{i}-r_{q}\sin\theta\;\mathbf{j}+z\;\mathbf{k}\right)}\\[5pt] \mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\iint {\frac{\sigma r_{q}\;dr_{q}\;d\theta}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}\left(-r_{q}\cos \theta\;\mathbf{i}-r_{q}\sin\theta\;\mathbf{j}+z\;\mathbf{k}\right)} \end{gather} \]

As the charge density σ is constant it is moved outside of the integral, and the integral of the sum is equal to the sum of the integrals

\[ \begin{gather} \mathbf{E}=\frac{\sigma}{4\pi \epsilon_{0}}\left(-\iint {\frac{r_{q}^{2}\cos \theta \;dr_{q}\;d\theta}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\;\mathbf{i}-\iint {\frac{r_{q}^{2}\sin\theta \;dr_{q}\;d\theta}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\;\mathbf{j}+z\iint {\frac{r_{q}\;dr_{q}\;d\theta}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\;\mathbf{k}\right) \end{gather} \]

The limits of integration will be b and a in dr_q, along the disk radius, 0 and 2π in dθ, a complete lap in the ring, the integral can be separated

\[ \begin{split} \mathbf{E}=\frac{\sigma}{4\pi \epsilon_{0}}\left(-\int_{b}^{a}{\frac{r_{q}^{2}\;dr_{q}}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\underbrace{\int_{0}^{{2\pi}}{\cos \theta \;d\theta}}_{0}\;\mathbf{i}-\int_{b}^{a}{\frac{r_{q}^{2}\;dr_{q}}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\underbrace{\int_{0}^{{2\pi}}{\operatorname{sen}\theta \;d\theta}}_{0}\;\mathbf{j}+\right.\\ \left.+z\int_{b}^{a}{\frac{r_{q}\;dr_{q}}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}}\int_{0}^{{2\pi}}{d\theta }\;\mathbf{k}\right) \end{split} \]

Integration of \( \displaystyle \int_{0}^{{2\pi}}\cos \theta \;d\theta \)

1st method

\[ \begin{align} \int_{0}^{{2\pi}}\cos \theta \;d\theta &=\left.\sin\theta\;\right|_{\;0}^{\;2\pi }=\sin2\pi-\sin0=\\ &=0-0=0 \end{align} \]

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).

Figure 3

Integration of \( \displaystyle \int_{0}^{{2\pi}}\sin\theta \;d\theta \)

1st method

\[ \begin{align} \int_{0}^{{2\pi}}\sin\theta \;d\theta &=\left.-\cos\theta \;\right|_{\;0}^{\;2\pi }=-(\cos 2\pi -\cos 0)=\\ &=-(1-1)=0 \end{align} \]

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).

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, dE_P, cancel. Only normal components to the plane dE_N contribute to the total electric field (Figure 5).

Figure 5

Integration of \( \displaystyle \int_{b}^{a}{\frac{r_{q}\;dr_{q}}{\left(r_{q}^{2}+z^{2}\right)^{\frac{3}{2}}}} \)

Changing the variable

\[ \begin{array}{l} u=r_{q}^{2}+z^{2}\\[5pt] \dfrac{du}{dr_{q}}=2r_{q}\Rightarrow dr_{q}=\dfrac{du}{2r_{q}} \end{array} \]

changing the limits of integration

for r_q = b
we have \( u=b^{2}+z^{2} \)

for r_q = a
we have \( u=a^{2}+z^{2} \)

\[ \begin{align} \int_{{b^{2}+z^{2}}}^{{a^{2}+z^{2}}}{\frac{r_{q}}{u^{\frac{3}{2}}}\;\frac{du}{2r_{q}}} &\Rightarrow\frac{1}{2}\int_{{b^{2}+z^{2}}}^{{a^{2}+z^{2}}}{\frac{1}{u^{\frac{3}{2}}}\;du}\Rightarrow\;\frac{1}{2}\left.\frac{u^{-{\frac{3}{2}+1}}}{-{\frac{3}{2}+1}}\;\right|_{\;b^{2}+z^{2}}^{\;a^{2}+z^{2}}\Rightarrow[5pt] &\Rightarrow\frac{1}{2}\left.\frac{u^{\frac{-3+2}{2}}}{\frac{-{3+2}}{2}}\;\right|_{\;b^{2}+z^{2}}^{\;a^{2}+z^{2}}\Rightarrow\frac{1}{2}\left.\frac{u^{-{\frac{1}{2}}}}{-{\frac{1}{2}}}\;\right|_{\;b^{2}+z^{2}}^{\;a^{2}+z^{2}}\Rightarrow\\[5pt] &\Rightarrow\left.-u^{-{\frac{1}{2}}}\;\right|_{\;b^{2}+z^{2}}^{\;a^{2}+z^{2}}\Rightarrow\left.-{\frac{1}{u^{\frac{1}{2}}}}\;\right|_{\;b^{2}+z^{2}}^{\;a^{2}+z^{2}}\Rightarrow\\[5pt] &\Rightarrow-\left(\frac{1}{\sqrt{a^{2}+z^{2}}}-\frac{1}{\sqrt{\;b^{2}+z^{2}}}\right)\Rightarrow\\[5pt] &\Rightarrow\frac{1}{\sqrt{b^{2}+z^{2}\;}}-\frac{1}{\sqrt{\;a^{2}+z^{2}\;}} \end{align} \]

Integration of \( \displaystyle \int_{0}^{{2\pi}}d\theta \)

\[ \begin{gather} \int_{0}^{{2\pi}}d\theta =\left.\theta \;\right|_{\;0}^{\;2\pi}=2\pi-0=2\pi \end{gather} \]

\[ \begin{gather} \mathbf{E}=\frac{\sigma}{4\pi \epsilon_{0}}\left[0\;\mathbf{i}-0\;\mathbf{j}+z\;\left(\frac{1}{\sqrt{b^{2}+z^{2}\;}}-\frac{1}{\sqrt{a^{2}+z^{2}\;}}\right)2\pi\;\mathbf{k}\right]\\[5pt] \mathbf{E}=\frac{\sigma}{\cancelto{2}{4}\cancel{\pi} \epsilon_{0}}\left[z\left(\frac{1}{\sqrt{b^{2}+z^{2}}}-\frac{1}{\sqrt{a^{2}+z^{2}\;}}\right)\cancel{2}\cancel{\pi}\;\mathbf{k}\right] \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathbf{E}=\frac{\sigma z}{2\epsilon_{0}}\left(\frac{1}{\sqrt{b^{2}+z^{2}\;}}-\frac{1}{\sqrt{a^{2}+z^{2}}}\;\right)\;\mathbf{k}} \end{gather} \]

Figure 6

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .