Solved Problem on Coulomb's Law and Electric Field
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A wire of length L carries a uniformly distributed charge Q. Determine the electric field vector at a point P of the line that contains the wire, x>L, is the coordinate of the outer point to the wire;
Problem data:
- Wire length: L;
- Wire charge: Q.
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, the problem is one-dimensional, all vectors are on the x-axis, in Figure 1 the elements were drawn separated to facilitate viewing.
\[
\mathbf{r}={\mathbf{r}}_{p}-{\mathbf{r}}_{q}
\]
From the geometry of the problem we choose Cartesian coordinates, we chose the origin of the system at the right end of the wire closest to point P, the vector rq is written as \( {\mathbf{r}}_{q}=-x_{q}\;\mathbf{i} \) and the rp vector as \( {\mathbf{r}}_{p}=x_{p}\;\mathbf{i} \), where xp is the distance to the desired point measured from the origin chosen, then the vector position will be
\[
\begin{gather}
\mathbf{r}=x_{p}\;\mathbf{i}-(-x_{q}\;\mathbf{i})\\
\mathbf{r}=(x_{p}+x_{q})\;\mathbf{i} \tag{I}
\end{gather}
\]
From expression (I), the magnitude of the position vector r will be
\[
\begin{gather}
r^{2}=(x_{p}+x_{q})^{2}\\
r=(x_{p}+x_{q}) \tag{II}
\end{gather}
\]
Solution
The electric field vector is given by
\[ \bbox[#99CCFF,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int{\frac{dq}{r^{2}}\;\frac{\mathbf{r}}{r}}}
\]
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int{\frac{dq}{r^{3}}\;\mathbf{r}} \tag{III}
\end{gather}
\]
Using the expression of the linear density of charge λ, we have the charge element dq
\[ \bbox[#99CCFF,10px]
{\lambda =\frac{dq}{ds}}
\]
\[
\begin{gather}
dq=\lambda \;ds \tag{IV}
\end{gather}
\]
where ds is an element of length of the wire
\[
\begin{gather}
ds=dx_{q} \tag{V}
\end{gather}
\]
substituting the expression (V) into expression (IV)
\[
\begin{gather}
dq=\lambda \;dx_{q} \tag{VI}
\end{gather}
\]
Substituting expressions (I), (II), and (VI) into expression (III)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\int {\frac{\lambda\;dx_{q}}{\left(x_{p}+x_{q}\right)^{\cancelto{2}{3}}}}\cancel{\left(x_{p}+x_{q}\right)}\;\mathbf{i}\\
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int {\frac{\lambda\;dx_{q}}{\left(x_{p}+x_{q}\right)^{2}}}\;\mathbf{i} \tag{VII}
\end{gather}
\]
As the charge density λ is constant, it is moved outside of the integral
\[
\mathbf{E}=\frac{\lambda}{4\pi \epsilon_{0}}\int{\frac{dx_{q}}{\left(x_{p}+x_{q}\right)^{2}}}\;\mathbf{i}
\]
The limits of integration will be 0 and L the length of the wire
\[
\mathbf{E}=\frac{\lambda}{4\pi \epsilon_{0}}\int_{0}^{L}{\frac{dx_{q}}{\left(x_{p}+x_{q}\right)^{2}}}\;\mathbf{i}
\]
Integration of \( \displaystyle \int_{0}^{L}{\frac{dx_{q}}{\left(x_{p}+x_{q}\right)^{2}}} \)
Changing the variable
for xq = 0
we have \( u=x_{p}-0\Rightarrow u=x_{p} \)
for xq = L
we have \( u=x_{p}+L \)
Changing the variable
\[
\begin{array}{l}
u=x_{p}+x_{q}\\
\dfrac{du}{dx_{q}}=1\Rightarrow dx_{q}=du
\end{array}
\]
changing the limits of integration
for xq = 0
we have \( u=x_{p}-0\Rightarrow u=x_{p} \)
for xq = L
we have \( u=x_{p}+L \)
\[
\begin{align}
\int_{x_{p}}^{{x_{p}+L}}{\frac{du}{u^{2}}} &\Rightarrow\int_{x_{p}}^{{x_{p}+L}}{u^{-2}du}\Rightarrow\left.\frac{u^{-2+1}}{-2+1}\;\right|_{\;x_{p}}^{\;x_{p}+L}\Rightarrow\\
&\Rightarrow\left.\frac{u^{-1}}{-1}\;\right|_{\;x_{p}}^{\;x_{p}+L}\Rightarrow-\left.\frac{1}{u}\;\right|_{\;x_{p}}^{\;x_{p}+L}\Rightarrow\\
&\Rightarrow-\frac{1}{x_{p}+L}-\left(-\frac{1}{x_{p}}\right)\Rightarrow \frac{1}{x_{p}}-\frac{1}{x_{p}-L}\Rightarrow\\
&\Rightarrow\frac{x_{p}+L-x_{p}}{x_{p}(x_{p}+L)}\Rightarrow\frac{L}{x_{p}(x_{p}+L)}
\end{align}
\]
\[
\begin{gather}
\mathbf{E}=\frac{\lambda}{4\pi \epsilon_{0}}\frac{L}{x_{p}\;(x_{p}+L)}\;\mathbf{i} \tag{VIII}
\end{gather}
\]
The linear density of charges can be written
\[
\begin{gather}
\lambda=\frac{Q}{L} \tag{IX}
\end{gather}
\]
substituting the expression (IX) into expression (VIII)
\[
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{Q}{\cancel{L}}\frac{\cancel{L}}{x_{p}(x_{p}+L)}\;\mathbf{i}
\]
\[ \bbox[#FFCCCC,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{Q}{x_{p}(x_{p}+L)}\;\mathbf{i}}
\]
and the magnitude of the electric field will be
\[ \bbox[#FFCCCC,10px]
{E=\frac{1}{4\pi \epsilon_{0}}\frac{Q}{x_{p}(x_{p}+L)}}
\]
Note: As the problem is one-dimensional, the vector value coincides with the scalar value
of the electric field.
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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 .