Solved Problem on Electric Field
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Two equal charges of the same sign are separated by a distance of 2d. Calculate the magnitude of the electric field at the points along the line joining the two charges. Check the solution for points far from the center of charges.
Problem diagram:
We choose a reference frame at the central point, 0, between the two charges, and x is the distance to the point P, where we want to calculate the electric field (Figure 1).
The distance from charge +q to point P is \( r=x-d \) (closer to point P), and the distance from the other charge +q to point P is \( r=x+d \) (far from point P). As one of the charges is closer to point P, it will produce a more intense field than the other charge.
Solution
The magnitude of the electric field of each charge is calculated by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{E=k_0\frac{q}{r^2}}
\end{gather}
\]
The resultant electric field will be given by
\[
\begin{gather}
E=E_1+E_2\\[5pt]
E=k_0\frac{q}{(x-d)^2}+k_0\frac{q}{(x+d)^2}\\[5pt]
E=k_0 q\left[\frac{(x+d)^2}{(x-d)^2}+\frac{(x-d)^2}{(x+d)^2}\right]
\end{gather}
\]
Terms in the numerator are in the form of Special Binomial Products
\[
\begin{gather}
(a+b)^2=a^2+2ab+b^2
\end{gather}
\]
\[
\begin{gather}
(a-b)^2=a^2-2ab+b^2
\end{gather}
\]
\[
\begin{gather}
E=k_0 q\left[\frac{x^2+2xd+d^2+x^2-2xd+d^2}{(x-d)^2(x+d)^2}\right]\\[5pt]
E=k_0 q\left[\frac{x^2+2xd+d^2+x^2-2xd+d^2}{(x-d)^2(x+d)^2}\right]
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{E=\frac{2k_0 q(x^2+d^2)}{(x-d)^2(x+d)^2}}
\end{gather}
\]
For points far from the center of the charges, x≫d, we can neglect the terms in d
in the denominator, and the term in d2 in the numerator and the solution will be
\[
\begin{gather}
E=\frac{2k_0 q\cancel{x^2}}{\cancel{x^2}x^2}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{E=\frac{2k_0 q}{x^2}}
\end{gather}
\]
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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 .