In (d), spherical coordinates, why only obtain radical component of E? Why not obtain E-theta or E-pi? want to know why not obtain E-Theta or E-Pi, but only obtain E-r
In (d), spherical coordinates, why only obtain radical component of E? Why not obtain E-theta or E-pi? want to know why not obtain E-Theta or E-Pi, but only obtain E-r
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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In (d), spherical coordinates, why only obtain radical component of E?
Why not obtain E-theta or E-pi?
want to know why not obtain E-Theta or E-Pi, but only obtain E-r
![|18 A charge Qọ located at the origin in free space produces a field for which E, =
1 kV/m at point P(-2, 1, – 1). (a) Find Qo. Find E at M(1, 6, 5) in (b) rectangular
coordinates; (c) cylindrical coordinates; (d) spherical coordinates.
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2252c1cf-c1b6-4684-8f54-52fea7ee7367%2F6e7f1e07-d29d-40ac-b2c4-7b8215049638%2Fm0p339e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:|18 A charge Qọ located at the origin in free space produces a field for which E, =
1 kV/m at point P(-2, 1, – 1). (a) Find Qo. Find E at M(1, 6, 5) in (b) rectangular
coordinates; (c) cylindrical coordinates; (d) spherical coordinates.
%3D
![d) Find E at M (1, 6, 5) in spherical coordinates: At M, r = V1+ 36+ 25 = 7.87, p = 80.54° (as
before), and 0
obtain only a radial component of EM
= cos-(5/7.87) = 50.58°. Now, since the charge is at the origin, we expect to
This will be:
·
E, = EM · a, = -30.11 sin 0 cos o – 180.63 sin 0 sin ø – 150.53 cos 0 = -237.1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2252c1cf-c1b6-4684-8f54-52fea7ee7367%2F6e7f1e07-d29d-40ac-b2c4-7b8215049638%2Fq5arx42_processed.jpeg&w=3840&q=75)
Transcribed Image Text:d) Find E at M (1, 6, 5) in spherical coordinates: At M, r = V1+ 36+ 25 = 7.87, p = 80.54° (as
before), and 0
obtain only a radial component of EM
= cos-(5/7.87) = 50.58°. Now, since the charge is at the origin, we expect to
This will be:
·
E, = EM · a, = -30.11 sin 0 cos o – 180.63 sin 0 sin ø – 150.53 cos 0 = -237.1
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