Solved Problem on Electric Field

Determine the electric field of a dipole at the points located on the perpendicular bisector of the dipole. Check the solution for points far from the center of the dipole.

Construction of the resultant electric field vector

On the bisector of the dipole, we choose any point P where we want to calculate the electric field. In the direction of the straight line that connects the charge +q to point P, we draw the vector \( \vec{E}_{\text{+}} \) pointing outward from the charge, q>0 (Figure 1).

Figure 1

In the direction of the line segment that connects the charge −q to point P, we draw the vector \( \vec{E}_{\text{-}} \) pointing inward to the charge, q<0 (Figure 2).

Figure 2

We draw a straight line through the end of vector \( \vec{E}_{\text{+}} \), parallel to vector \( \vec{E}_{\text{-}} \) (Figure 3).

Figure 3

We draw a straight line through the end of vector \( \vec{E}_{\text{-}} \), parallel to vector \( \vec{E}_{\text{+}} \) (Figure 4).

Figure 4

From point P to the intersection of the straight lines, we have the resultant vector \( \vec{E} \). The angle θ that the components of the electric field, \( \vec{E}_{\text{+}} \), and \( \vec{E}_{\text{-}} \), make with the resultant vector \( \vec{E} \) is the same angle that the segment r makes with the vertical segment d (Figure 5).

Figure 5

Solution

The magnitude of the electric field of each charge is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {E=k_{0}\frac{q}{r^{2}}} \tag{I} \end{gather} \]

The resultant electric field will be given by

\[ \begin{gather} \vec{E}=\vec{E}_{1}+\vec{E}_{2} \end{gather} \]

as the charges have the same value in magnitude, \( E_{1}=E_{2} \)

\[ \begin{gather} E=E_{1}\cos \theta+E_{2}\cos \theta\\[5pt] E=2E_{1}\cos\theta \tag{II} \end{gather} \]

The cosine of θ is obtained from r and d

\[ \begin{gather} \cos \theta=\frac{d}{r} \tag{III} \end{gather} \]

the segment r is obtained using the Pythagorean Theorem

\[ \begin{gather} r^{2}=d^{2}+x^{2}\\[5pt] r=\sqrt{d^{2}+x^{2}\;} \tag{IV} \end{gather} \]

substituting equation (IV) into equation (III)

\[ \begin{gather} \cos \theta=\frac{d}{\sqrt{d^{2}+x^{2}\;}} \tag{V} \end{gather} \]

Substituting equations (I) and (V) into equation (II)

\[ \begin{gather} E=2k_{0}\frac{q}{\left(\sqrt{d^{2}+x^{2}\;}\right)^{2}}\frac{d}{\sqrt{d^{2}+x^{2}\;}}\\[5pt] E=2k_{0}\frac{q}{\left(d^{2}+x^{2}\right)}\frac{d}{\left(d^{2}+x^{2}\right)^{\frac{1}{2}}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {E=\frac{2k_{0}qd}{\left(d^{2}+x^{2}\right)^{\frac{3}{2}}}} \end{gather} \]

For points far from the center of the dipole we have, x≫d, we can neglect the term in d² in the denominator and the solution will be

\[ \begin{gather} E=\frac{2k_{0}qd}{x^{{\cancel{2}}\times{\frac{3}{\cancel{2}}}}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {E=\frac{2k_{0}qd}{x^{3}}} \end{gather} \]

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