Solved Problem on Coulomb's Law and Electric Field

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 from the charges.

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 \( \mathbf r=x\;\mathbf i \). The vector r₁ goes from the origin to the charge +q (on the left in Figure 1), it is given by \( \mathbf r_1=-d\;\mathbf i \). The vector r−r₁ goes from the charge to point P, it is given by \( \mathbf r-\mathbf r_1=x\;\mathbf i-\left(-d\;\mathbf i\right)=\left(x+d\right)\;\mathbf i \).

Figure 1

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 \( \mathbf r_2=d\mathbf i \). The vector r−r₂ goes from the charge to the point P, and is given by \( \mathbf r-\mathbf r_2=\left(x-d\right)\;\mathbf i \), (Figure 2).

Figura 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 \( \left|\mathbf r-\mathbf r_1\right|=\sqrt{\left(x+d\right)^2\;}=x+d \), \( \left|\mathbf r-\mathbf r_1\right|^2=\left(x+d\right)^2 \), \( \left|\mathbf r-\mathbf r_2\right|=\sqrt{\left(x-d\right)^2\;}=x-d \). and \( \left|\mathbf r-\mathbf r_2\right|^2=\left(x-d\right)^2 \)

\[ \begin{gather} \mathbf E=\frac{1}{4\pi \epsilon_0}\left\{\frac{q}{\left(x+d\right)^2}\;\frac{\cancel{\left(x+d\right)}\;\mathbf i}{\cancel{\left(x+d\right)}}+\frac{q}{\left(x-d\right)^2}\;\frac{\cancel{\left(x-d\right)}\;\mathbf i}{\cancel{\left(x-d\right)}}\right\}\\[5pt] \mathbf E=\frac{q}{4\pi\epsilon_0}\left\{\frac{1}{\left(x+d\right)^2}+\frac{1}{\left(x-d\right)^2}\right\}\;\mathbf i\\[5pt] \mathbf E=\frac{q}{4\pi\epsilon_0}\left\{\frac{\left(x-d\right)^2+\left(x+d\right)^2}{\left(x+d\right)^2\left(x-d\right)^2}\right\}\;\mathbf i\\[5pt] \mathbf E=\frac{q}{4\pi\epsilon_0}\left\{\frac{x^2-2xd+d^2+x^2+2xd+d^2}{\left(x+d\right)^2\left(x-d\right)^2}\right\}\;\mathbf i\\[5pt] \mathbf E=\frac{1}{4\pi\epsilon_0}\frac{2q\left(x^2+d^2\right)}{\left(x+d\right)^2\left(x-d\right)^2}\;\mathbf i \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathbf E=\frac{1}{2\pi \epsilon_0}\frac{q\left(x^2+d^2\right)}{\left(x+d\right)^2\left(x-d\right)^2}\;\mathbf i} \end{gather} \]

Note 1: In a one-dimensional problem, the vector solution is the same as the scalar solution.

For points far from the center we have, y≫d, we can neglect the terms in d in the denominator and the term in d² in the numerator, and the solution will be

\[ \begin{gather} \mathbf E=\frac{1}{2\pi \epsilon_0}\frac{q\cancel{x^2}}{x^\cancelto{2}{4}}\;\mathbf i \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathbf E=\frac{1}{2\pi \epsilon_0}\frac{q}{x^2}\;\mathbf i} \end{gather} \]

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