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What i couldn't understand from this is whe he was getting the value for cosa he got 5/13 at which 13 is the hypotenuse but how did he get the value 5? If possible can you elaborate or explain that?

What i couldn't understand from this is whe he was getting the value for cosa he got 5/13 at which 13 is the hypotenuse but how did he get the value 5? If possible can you elaborate or explain that?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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What i couldn't understand from this is whe he was getting the value for cosa he got 5/13 at which 13 is the hypotenuse but how did he get the value 5? If possible can you elaborate or explain that?
21.5 Electric-Field Calculations 705 21.22 Electric field at three points, a, b, and e, set up by charges q, and 92, which form an electric dipole. (c) Figure 21.22 shows the directions of E, and E2 at c. Both vectors have the same x-component: Eia = Em = E,, cos œ = (6.39 × 10' N/C)(÷) = 2.46 × 10° N/C From symmetry, E, and E,, are equal and opposite, so their sum is zero. Hence E = 2(2.46 × 10 N/C)i = (4.9 × 10' N/C)i EVALUATE: We can also find E using Eq (21 7) for the field of a point charge. The displacement vector Fị from q to point e is r Cosai + sinaj. Hence the unit veetor that points from 130 cm 13.0 cm 41 to point c unit vector that points from q. to point e has the opposite x- component but We can now use Eq. (21.7) to write the fields E and E, at e in vector form rto c is the same for both charges. rr= CONal + sinaj. By symmetry. the same v-component , = -cosaf + sinaj. their sum. Since q1 4i and the distance 6.0 E = È +E˸= 42 4.0 ta) Ata. E, and E, até both directed to the righL SO (2 cusai) 12 10 (0.13 m Ē, = L i + Ei = (9.8 × 10 NC = 2(9.0 X 10 N mC (b) At , E, is directed to the left and E, is directed to the tight, so = (49 x 10 N/C)
Transcribed Image Text:21.5 Electric-Field Calculations 705 21.22 Electric field at three points, a, b, and e, set up by charges q, and 92, which form an electric dipole. (c) Figure 21.22 shows the directions of E, and E2 at c. Both vectors have the same x-component: Eia = Em = E,, cos œ = (6.39 × 10' N/C)(÷) = 2.46 × 10° N/C From symmetry, E, and E,, are equal and opposite, so their sum is zero. Hence E = 2(2.46 × 10 N/C)i = (4.9 × 10' N/C)i EVALUATE: We can also find E using Eq (21 7) for the field of a point charge. The displacement vector Fị from q to point e is r Cosai + sinaj. Hence the unit veetor that points from 130 cm 13.0 cm 41 to point c unit vector that points from q. to point e has the opposite x- component but We can now use Eq. (21.7) to write the fields E and E, at e in vector form rto c is the same for both charges. rr= CONal + sinaj. By symmetry. the same v-component , = -cosaf + sinaj. their sum. Since q1 4i and the distance 6.0 E = È +E˸= 42 4.0 ta) Ata. E, and E, até both directed to the righL SO (2 cusai) 12 10 (0.13 m Ē, = L i + Ei = (9.8 × 10 NC = 2(9.0 X 10 N mC (b) At , E, is directed to the left and E, is directed to the tight, so = (49 x 10 N/C)
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