Solved Problem on Electric Field

Two equal charges of the same sign are separated by a distance 2d. Calculate the magnitude of the electric field 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 charges.

Construction of the resultant electric field vector

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

Figure 1

In the direction of the line segment that connects the other charge +q to point P, we draw the vector \( {\vec E}_{2} \) pointing outward from the charge, q>0 (Figure 2).

Figure 2

We draw a straight line through the end of vector \( {\vec E}_{1} \), parallel to vector \( {\vec E}_{2} \) (Figure 3).

Figure 3

We draw a straight line through the end of vector \( {\vec E}_{2} \), parallel to vector \( {\vec E}_{1} \) (Figure 4).

Figure 4

From point P to the intersection of the lines, we have the resultant vector \( \vec{E} \), and θ is the angle between the electric field vector \( {\vec E}_{2} \) (or the vector \( {\vec E}_{1} \) ) and an auxliliary line parallel to a line that connects the two charges. The angle θ that the electric field vectors, \( {\vec E}_{1} \) and \( {\vec E}_{2} \), make with the auxiliary line, is the same angle that the segment r makes with the vertical segment d (Figure 5).

Figure 5

Note 1: This system does not represent an electric dipole, a dipole is formed by charges of the same value and opposite signs, in this case we have charges of the same sign.

Note 2: Instead of using the angle θ between the segment r and the vertical segment d between the two charges, we could use the angle between the segment r and the segment x (Figure 6). The sum of the interior angles of a triangle is equal to 180°

\[ \begin{gather} 180°=90°+\theta+\alpha\\[5pt] \alpha=90°\theta \end{gather} \]

Figure 6

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, \( E_{1}=E_{2} \)

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

Note: Using the cos 90°−θ obtained above, the electric field will be

\[ \begin{gather} E=2E_{1}\;\cos (90°-\theta) \end{gather} \]

the difference formula for cosine is given by

\[ \begin{gather} \cos (a-b)=\cos a\cos b+\sin a\sin b \\[10pt] \cos (90°-\theta)=\underbrace{\cos90°}_{0}\cos\theta+\underbrace{\sin 90°}_{1}\sin\theta\\[5pt] \cos(90°-\theta)=\sin \theta \end{gather} \]

The sine of θ is obtained from r and x

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

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)

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

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

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

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

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

\[ \begin{gather} E=\frac{2k_{0}qx}{x^{{\cancel{2}}\times{\frac{3}{\cancel{2}}}}} \\[5pt] E=\frac{2k_{0}q\cancel{x}}{x^{\cancelto{2}{3}}} \end{gather} \]

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

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