Solved Problem on Electric Field

Determine the electric field of a dipole at the points along the straight line joining the two charges of the electric dipole. Check the solution for points far from the center of the dipole.

Problem diagram:

We choose a reference frame at the central point, 0, between the two charges of the dipole, and x is the distance from the center of the dipole to the point P, where we want to calculate the electric field (Figure 1).

Figure 1

The distance from charge +q to point P isé \( r=x-d \) and the distance from charge −q to point P is \( r=x+d \). Since charge +q is closer to point P, it will produce a stronger field than charge −q.

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

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

Terms in the numerator are in the form of Special Binomial Products

\[ \begin{gather} (a+b)^{2}=a^{2}+2ab+b^{2} \end{gather} \]

\[ \begin{gather} (a-b)^{2}=a^{2}-2ab+b^{2} \end{gather} \]

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

\[ \begin{gather} \bbox[#FFCCCC,10px] {E=\frac{4k_{0}qxd}{(x-d)^{2}(x+d)^{2}}} \end{gather} \]

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

\[ \begin{gather} E=\frac{4k_{0}q\cancel{x}d}{x^{\cancel{2}}x^{2}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {E=\frac{4k_{0}qd}{x^{3}}} \end{gather} \]

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