Question 3 Two, hollow, spherical conductors, are placed concentrically in free space. The inner sphere has radius Ri and the outer Ro. The inner sphere has a charge Q, the outer sphere has a charge -Q, see Figure 3.
Question 3 Two, hollow, spherical conductors, are placed concentrically in free space. The inner sphere has radius Ri and the outer Ro. The inner sphere has a charge Q, the outer sphere has a charge -Q, see Figure 3.
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Please could I have the written out step by step solution for this question.
![Question 3
Two, hollow, spherical conductors, are placed concentrically in free space. The inner sphere
has radius Ri and the outer Ro. The inner sphere has a charge Q, the outer sphere has a
charge -Q, see Figure 3.
-Q
R.
E
E
Figure 3
(a) Use Gauss's law to show that, between the two spheres, the electric field is
E =
4ne,r²
and zero everywhere else. State all assumptions that you make.
Hint: you should set the Gaussian surface to be the surface of a sphere of radius r,
sharing the same centre as the charged spheres. Remember that if the flux density is
perpendicular to the Gaussian surface and constant over its surface then Gauss' law
reduces to
D dA = DA = Q
where A is the Gaussian surface's area.
(b) Use the result in equation (1) and the definition of potential difference,
.------
.------](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F756bb6af-597e-47da-9902-dc67cb35fa89%2F36c534b4-d3ce-4149-93fb-cc9747dd8ed4%2Fkjleuxn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 3
Two, hollow, spherical conductors, are placed concentrically in free space. The inner sphere
has radius Ri and the outer Ro. The inner sphere has a charge Q, the outer sphere has a
charge -Q, see Figure 3.
-Q
R.
E
E
Figure 3
(a) Use Gauss's law to show that, between the two spheres, the electric field is
E =
4ne,r²
and zero everywhere else. State all assumptions that you make.
Hint: you should set the Gaussian surface to be the surface of a sphere of radius r,
sharing the same centre as the charged spheres. Remember that if the flux density is
perpendicular to the Gaussian surface and constant over its surface then Gauss' law
reduces to
D dA = DA = Q
where A is the Gaussian surface's area.
(b) Use the result in equation (1) and the definition of potential difference,
.------
.------
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