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Example 1.1 Three Equal Point Charges at Triangle Vertices Three small charged bodies of charge Q are placed at three vertices of an equilateral triangle with sides a, in air. The bodies can be considered as point charges. Find the direction and magnitude of the electric force on each of the charges. Solution Even with no computation whatsoever, we can conclude from the symmetry of this problem that the resultant forces on the charges, Fel, Fe2, and Fe3, all have the same magnitude and are positioned in the plane of the triangle as indicated in Fig. 1.3(a). Let us compute the resultant force on the lower right charge - charge 3. Using the principle of superposition, this force represents the vector sum of partial forces due to charges 1 and 2, respectively, that is [Fig. 1.3(b)], Fe3 = Fe 13 Fe 23 (vector superposition) . From Coulomb's law, Eq. (1.1), magnitudes of the individual partial forces are given by Fe 13 = Fe23 = Q2 4* €0a2 (1.8) (1.9) From Coulomb's law, Eq. (1.1), magnitudes of the individual partial forces are given by Fe 13 = Fe3= 2 (Fe 13 cos a) Fe 23 = = and both forces are repulsive. We note that the vector Fe3 is positioned along the symmetry line between charges 1 and 2, i.e., between vectors Fe13 and Fe23, and it makes the angle a = π/6 with both vectors. The magnitude of the resultant vector is therefore twice the projection of any of the partial vectors on the symmetry line, which yields Q² 4π €0a⁹ 2 Fe 13 e 13 V3 2 = Fe 13 √ 13 = (1.9) √30² 4 €0a² (1.10)

Example 1.1 Three Equal Point Charges at Triangle Vertices Three small charged bodies of charge Q are placed at three vertices of an equilateral triangle with sides a, in air. The bodies can be considered as point charges. Find the direction and magnitude of the electric force on each of the charges. Solution Even with no computation whatsoever, we can conclude from the symmetry of this problem that the resultant forces on the charges, Fel, Fe2, and Fe3, all have the same magnitude and are positioned in the plane of the triangle as indicated in Fig. 1.3(a). Let us compute the resultant force on the lower right charge - charge 3. Using the principle of superposition, this force represents the vector sum of partial forces due to charges 1 and 2, respectively, that is [Fig. 1.3(b)], Fe3 = Fe 13 Fe 23 (vector superposition) . From Coulomb's law, Eq. (1.1), magnitudes of the individual partial forces are given by Fe 13 = Fe23 = Q2 4* €0a2 (1.8) (1.9) From Coulomb's law, Eq. (1.1), magnitudes of the individual partial forces are given by Fe 13 = Fe3= 2 (Fe 13 cos a) Fe 23 = = and both forces are repulsive. We note that the vector Fe3 is positioned along the symmetry line between charges 1 and 2, i.e., between vectors Fe13 and Fe23, and it makes the angle a = π/6 with both vectors. The magnitude of the resultant vector is therefore twice the projection of any of the partial vectors on the symmetry line, which yields Q² 4π €0a⁹ 2 Fe 13 e 13 V3 2 = Fe 13 √ 13 = (1.9) √30² 4 €0a² (1.10)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Three unequal charges in a triangle. Repeat Example 1.1, but assuming that one of the three charges in Fig. 1.3(a) amounts to (a) 3Q and (b) –3Q, respectively.

 

Example 1.1 Three Equal Point Charges at Triangle Vertices Three small charged bodies of charge Q are placed at three vertices of an equilateral triangle with sides a, in air. The bodies can be considered as point charges. Find the direction and magnitude of the electric force on each of the charges. Solution Even with no computation whatsoever, we can conclude from the symmetry of this problem that the resultant forces on the charges, Fel, Fe2, and Fe3, all have the same magnitude and are positioned in the plane of the triangle as indicated in Fig. 1.3(a). Let us compute the resultant force on the lower right charge - charge 3. Using the principle of superposition, this force represents the vector sum of partial forces due to charges 1 and 2, respectively, that is [Fig. 1.3(b)], Fe3 = Fe 13 Fe 23 (vector superposition) . From Coulomb's law, Eq. (1.1), magnitudes of the individual partial forces are given by Fe 13 = Fe23 = Q2 4* €0a2 (1.8) (1.9)
Transcribed Image Text:Example 1.1 Three Equal Point Charges at Triangle Vertices Three small charged bodies of charge Q are placed at three vertices of an equilateral triangle with sides a, in air. The bodies can be considered as point charges. Find the direction and magnitude of the electric force on each of the charges. Solution Even with no computation whatsoever, we can conclude from the symmetry of this problem that the resultant forces on the charges, Fel, Fe2, and Fe3, all have the same magnitude and are positioned in the plane of the triangle as indicated in Fig. 1.3(a). Let us compute the resultant force on the lower right charge - charge 3. Using the principle of superposition, this force represents the vector sum of partial forces due to charges 1 and 2, respectively, that is [Fig. 1.3(b)], Fe3 = Fe 13 Fe 23 (vector superposition) . From Coulomb's law, Eq. (1.1), magnitudes of the individual partial forces are given by Fe 13 = Fe23 = Q2 4* €0a2 (1.8) (1.9)
From Coulomb's law, Eq. (1.1), magnitudes of the individual partial forces are given by Fe 13 = Fe3= 2 (Fe 13 cos a) Fe 23 = = and both forces are repulsive. We note that the vector Fe3 is positioned along the symmetry line between charges 1 and 2, i.e., between vectors Fe13 and Fe23, and it makes the angle a = π/6 with both vectors. The magnitude of the resultant vector is therefore twice the projection of any of the partial vectors on the symmetry line, which yields Q² 4π €0a⁹ 2 Fe 13 e 13 V3 2 = Fe 13 √ 13 = (1.9) √30² 4 €0a² (1.10)
Transcribed Image Text:From Coulomb's law, Eq. (1.1), magnitudes of the individual partial forces are given by Fe 13 = Fe3= 2 (Fe 13 cos a) Fe 23 = = and both forces are repulsive. We note that the vector Fe3 is positioned along the symmetry line between charges 1 and 2, i.e., between vectors Fe13 and Fe23, and it makes the angle a = π/6 with both vectors. The magnitude of the resultant vector is therefore twice the projection of any of the partial vectors on the symmetry line, which yields Q² 4π €0a⁹ 2 Fe 13 e 13 V3 2 = Fe 13 √ 13 = (1.9) √30² 4 €0a² (1.10)
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