Solved Problem on Coulomb's Law and Electric Field
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Two equal charges of the same sign are separated by a distance of 2d. Calculate the electric field vector at the points along the perpendicular bisector of the line joining the two charges. Check the solution for points far away from the center of the dipole.
Problem diagram:
The vector r locates the point P, where we want to calculate the electric field relative
to the origin, and is written as
\( \mathbf{r}=d\;\mathbf{i}+y\;\mathbf{j} \).
The vector r1 goes from the origin to the charge +q (on the left in Figure 1),
as the charge is located at the origin this vector is equal to zero,
\( \mathbf{r}_{1}=\mathbf{0} \).
The vector r−r1 goes from the charge to point P, in this case it
coincides with the vector r, it is given by
\( \mathbf{r}-\mathbf{r}_{1}=d\;\mathbf{i}+y\;\mathbf{j}-\mathbf{0}=d\;\mathbf{i}+y\;\mathbf{j} \).
The vector r is the same as in the previous situation. The vector r2 goes from
the origin to the second charge +q, and is given by
\( \mathbf{r}_{2}=2d\mathbf{i} \).
The vector r−r2 goes from the charge to the point P, and is given
by
\( \mathbf{r}-\mathbf{r}_{2}=d\;\mathbf{i}+y\;\mathbf{j}-2d\mathbf{i}=-d\;\mathbf{i}+y\;\mathbf{j} \),
(Figure 2).
Solution
The electric field vector of a discrete system of charges is calculated by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\;\sum_{i=1}^{n}\;\frac{q_{i}}{\left|\mathbf{r}-\mathbf{r}_{i}\right|^{2}}\;\frac{\mathbf{r}-\mathbf{r}_{i}}{\left|\mathbf{r}-\mathbf{r}_{i}\right|}}
\end{gather}
\]
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\left\{\frac{q_{1}}{\left|\mathbf{r}-\mathbf{r}_{1}\right|^{2}}\;\frac{\mathbf{r}-\mathbf{r}_{1}}{\left|\mathbf{r}-\mathbf{r}_{1}\right|}+\frac{q_{2}}{\left|\mathbf{r}-\mathbf{r}_{2}\right|^{2}}\;\frac{\mathbf{r}-\mathbf{r}_{2}}{\left|\mathbf{r}-\mathbf{r}_{2}\right|}\right\}
\end{gather}
\]
The denominators of the above equation are written as
\( \left|\mathbf{r}-\mathbf{r}_{1}\right|=\sqrt{d^{2}+y^{2}\;} \),
\( \left|\mathbf{r}-\mathbf{r}_{1}\right|^{2}=d^{2}+y^{2} \),
\( \left|\mathbf{r}-\mathbf{r}_{2}\right|=\sqrt{(-d)^{2}+y^{2}\;}=\sqrt{d^{2}+y^{2}\;} \).
and
\( \left|\mathbf{r}-\mathbf{r}_{2}\right|^{2}=d^{2}+y^{2} \)
\[
\begin{gather}
\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\left\{\frac{q}{\left(d^{2}+y^{2}\right)^{1/2}}\;\frac{\left(d\;\mathbf{i}+y\;\mathbf{j}\right)}{\left(d^{2}+y^{2}\right)}+\frac{q}{\left(d^{2}+y^{2}\right)^{1/2}}\;\frac{\left(-d\;\mathbf{i}+y\;\mathbf{j}\right)}{\left(d^{2}+y^{2}\right)}\right\}\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{\left(d^{2}+y^{2}\right)^{3/2}}\left[d\;\mathbf{i}+y\;\mathbf{j}-d\;\mathbf{i}-y\;\mathbf{j}\right]\\[5pt]
\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\frac{2yq}{\left(d^{2}+y^2\right)^{3/2}}\;\mathbf{j}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{2qy}{\left(d^{2}+y^{2}\right)^{3/2}}\;\mathbf{j}}
\end{gather}
\]
For points far from the center of the dipole we have, y≫d, we can neglect the term in
d2 in the denominator, and the solution will be
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\mathbf{E}=\frac{1}{4\pi \epsilon_{0}}\frac{2q}{y^{2}}\;\mathbf{j}}
\end{gather}
\]
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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 .