Solved Problem on Electric Field

Four positive charges equal to Q, are placed on a horizontal plane in the vertices of a square of side d.
a) Find the magnitude of the electric field at a point P above the center of the square at a distance equal to d. Assume that the Coulomb constant is equal to k₀.
b) If a charge Q < 0 is placed in P, find the magnitude of the electric force will act on this charge.

Problem data:

Electric charges: +q;
Distance between the charges: d;
Coulomb constant: k₀.

Problem diagram:

As the charges are positive they create an electric field away from the point P, and as the values of the charges are equal, the magnitude of the electric fielde E due to each charge will be the same (Figure 1).

Figure 1

Solution

a) Looking to a vertical plane passing through one of the charges at the base and point P (Figure 2), we have that the magnitude of the electric field is

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

This electric field must be projected in the directions parallel to the plane of the charges E_P, and normal to the plane E_N, drawing the electric field in a system of coordinates xy and calculating its components, we have (Figure 3)

Figure 2

\[ \begin{gather} E_{P}=E\sin\theta \tag{II-a} \end{gather} \]

\[ \begin{gather} E_{N}=E\cos \theta \tag{II-b} \end{gather} \]

Figure 3

where the angle θ, between the vector of the electric field \( \vec{E} \) and the normal component \( {\vec{E}}_{N} \) in the plane, is the same angle between the distance r of the charge to the point P and the height d of the center of the square to the point (they are vertically opposite angles).

Electric field parallel to the plane (\( {\vec{E}}_{P} \)):

By symmetry of the problem, each charge of the square will create an electric field of the same magnitude in P, so we have four parallel components \( {\vec{E}}_{P} \). Looking from above (Figure 4-B), we see that the electric field created by the charges cancels out, or by the Polygon Method for the sum of vectors (Figure 4-C), we have these four electric field vectors sum is zero.

\[ \begin{gather} {\vec{E}}_{P}+{\vec{E}}_{P}+{\vec{E}}_{P}+{\vec{E}}_{P}=0\\ E_{P}-E_{P}+E_{P}-E_{P}=0 \end{gather} \]

Figure 4

Electric field normal to the plane (\( {\vec{E}}_{N} \)):

To find the component normal to the plane \( {\vec{E}}_{N} \), we must find the cosine of the angle θ as a function of the distance d, between the center of the base and the point P. The cosine of θ is calculated by

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

The diagonal length of the base can be found using the Pythagorean Theorem (Figure 5)

\[ \begin{gather} h^{2}=d^{2}+d^{2}\\ h^{2}=2d^{2}\\ h=\sqrt{2d^{2}\;}\\ h=d\sqrt{2\;} \tag{IV} \end{gather} \]

Figure 5

Diagonal h is divided into two segments of sizes \( \frac{d\sqrt{2\;}}{2} \) (Figura 6), using the Pythagorean Theorem, we can determine r as a function of d

\[ \begin{gather} r^{2}=d^{2}+\left(\frac{d\sqrt{2\;}}{2}\right)^{2}\\ r^{2}=d^{2}+\frac{2d^{2}}{4}\\ r^{2}=d^{2}+\frac{d^{2}}{2} \end{gather} \]

Figure 6

multiplying and dividing by 2 the first term on the right-hand side of the equation

\[ \begin{gather} r^{2}=\frac{2}{2}\times d^{2}+\frac{d^{2}}{2}\\ r^{2}=\frac{2d^{2}+d^{2}}{2}\\ r^{2}=\frac{3d^{2}}{2}\\ r=\sqrt{\frac{3d^{2}}{2}\;}\\ r=d\sqrt{\frac{3}{2}\;}\\ r=d\frac{\sqrt{3\;}}{\sqrt{2\;}} \end{gather} \]

multiplying the numerator and the denominator by \( \sqrt{2\;} \), we have

\[ \begin{gather} r=d\frac{\sqrt{3\;}}{\sqrt{2\;}}\times \frac{\sqrt{2\;}}{\sqrt{2\;}}\\ r=d\frac{\sqrt{6\;}}{2} \tag{V} \end{gather} \]

substituting the expression (V) into (III) the cosine of θ is

\[ \begin{gather} \cos \theta =\frac{d}{\dfrac{d\sqrt{6\;}}{2}}\\[5pt] \cos\theta =\frac{2}{\sqrt{6\;}} \end{gather} \]

multiplying the numerator and the denominator by \( \sqrt{6\;} \), we have

\[ \begin{gather} \cos \theta=\frac{2}{\sqrt{6\;}}\times \frac{\sqrt{6\;}}{\sqrt{6\;}}\\ \cos \theta=\frac{2\sqrt{6\;}}{6} \end{gather} \]

Substituting the expression (I) into (II-b) the normal component of the electric field to the in the point P is

\[ \begin{gather} E_{N}=k_{0}\frac{q}{r^{2}}\cos \theta \tag{VI} \end{gather} \]

substituting the expression (V) and the cosine found above (VI), we obtain

\[ \begin{gather} E_{N}=k_{0}\frac{q}{\left(d\dfrac{\sqrt{6\;}}{2}\right)^{2}}\times \frac{2\sqrt{6\;}}{6}\\[5pt] E_{N}=k_{0}\frac{q}{d^{2}\dfrac{6}{4}}\times \frac{2\sqrt{6\;}}{6}\\[5pt] E_{N}=k_{0}\frac{q}{d^{2}}\frac{2\sqrt{6\;}}{6}\times \frac{4}{6}\\[5pt] E_{N}=k_{0}\frac{q}{d^{2}}\frac{8\sqrt{6\;}}{36}\\[5pt] E_{N}=\frac{2\sqrt{6\;}}{9}k_{0}\frac{q}{d^{2}} \end{gather} \]

Figure 7

By symmetry the magnitude of the electric field at point P created by the other three charges of the base is the same, then the resultant magnitude of the electric field in P will be (Figure 7)

\[ E_{R}=4E_{N} \]

\[ \bbox[#FFCCCC,10px] {E_{R}=\frac{8\sqrt{6\;}}{9}k_{0}\frac{q}{d^{2}}} \]

b) The magnitude of electric force is given by

\[ \bbox[#99CCFF,10px] {F_{E}=qE} \]

using the result of the previous item and the charge given, Q < 0, the the electric force on the charge at the point P

\[ F_{E}=-Q\frac{8\sqrt{6\;}}{9}k_{0}\frac{q}{d^{2}} \]

\[ \bbox[#FFCCCC,10px] {F_{E}=-{\frac{8\sqrt{6\;}}{9}}k_{0}\frac{Qq}{d^{2}}} \]

Figure 8

the negative sign indicates that the electric force has a direction opposite to the electric field (Figure 8).

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .