Solved Problem on Electric Field
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Two point charges of 5×10−6 C and 3×10−6 C, are placed in two vertices of an equilateral triangle of sides equal to 1.2 m. Calculate the magnitude of the electric field in the third vertex, assuming they are in the vacuum.


Problem data:
  • Charge 1:    q1 = 5×10−6 C;
  • Charge 2:    q2 = 3×10−6;
  • Distance between the charges:    d = 1,2 m;
  • Coulomb constant:    \( k_{0}=9\times 10^{9}\;\frac{\text{N.m}^{2}}{\text{C}^{2}} \).
Construção do vetor campo elétrico resultante

Charges q1 and q2 are placed in vertices A and B of the triangle. Considering the charge q1 of the highest value 5×10−6 C, we draw at point C the vector \( {\vec{E}}_{1} \), in the direction of the segment \( \overline{AC} \), with direction pointing away from the charge, q > 0, the highest value load creates a field more intense (Figure 1).
Figure 1

At point C we draw the vector \( {\vec{E}}_{2} \), in the direction of the segment \( \overline{BC} \), with direction away and smaller size, the charge q2 is of smaller value, 3×10−6 C, and creates a less intense field (Figure 2).
Figure 2

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

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

From vertex C to the intersection of the lines, we have the resultant vector \( \vec{E} \), α is the angle between the electric field vectors \( {\vec{E}}_{1} \) and \( {\vec{E}}_{2} \). As the triangle is equilateral internal angles are equal to β = 60°, as well as angles α and β are vertically opposite angles, the angle α is also 60° (Figure 5).
Figure 5

Solution

The resultant electric field is given by
\[ \vec{E}={\vec{E}}_{1}+{\vec{E}}_{2} \]
the magnitude can be calculated using the Law of Cosines
\[ \begin{gather} \bbox[#99CCFF,10px] {E^{2}=E_{1}^{2}+E_{3}^{2}+2E_{1}E_{2}\cos \alpha} \tag{I} \end{gather} \]
The magnitude of the electric field of each charge is calculated by
\[ \bbox[#99CCFF,10px] {E=k_{0}\frac{q}{r^{2}}} \]
\[ \begin{gather} E_{1}=k_{0}\frac{q_{1}}{r_{1}^{2}}\\ E_{1}=9\times 10^{9}\times \frac{5\times 10^{-6}}{(1.2)^{2}}\\ E_{1}=\frac{4.5\times 10^{4}}{1.44}\\ E_{1}\simeq 3.1\times 10^{4}\;\frac{\text{N}}{\text{C}} \tag{II} \end{gather} \]
\[ \begin{gather} E_{2}=k_{0}\frac{q_{2}}{r_{2}^{2}}\\ E_{2}=9\times 10^{9}\times \frac{3\times 10^{-6}}{(1.2)^{2}}\\ E_{2}=\frac{2.7\times 10^{4}}{1.44}\\ E_{2}\simeq 1.9\times 10^{4}\;\frac{\text{N}}{\text{C}} \tag{III} \end{gather} \]
substituting expressions (II) and (III) into (I), we have
\[ \begin{gather} E^{2}=(3.1\times 10^{4})^{2}+(1.9\times 10^{4})^{2}+2\times 3.1\times 10^{4}\times 1.9\times 10^{4}\cos60°\\ E^{2}=9.6\times 10^{8}+3.6\times 10^{8}+\cancel{2}\times 5.9\times 10^{8}\times \frac{1}{\cancel{2}}\\ E^{2}=(9.6+3.6+5.9)\times 10^{8}\\ E^{2}=19.1\times 10^{8}\\ E=\sqrt{19.1\times 10^{8}\;} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {E\simeq 4.4\times 10^{4}\;\frac{\text{N}}{\text{C}}} \]
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