College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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A hollow sphere of radui R has a uniform nagative surface charge density -\sigma on its surface and a positive point charge +Q at its center. The charge Q is greater than the absolute magnitude of the total charge on the surface. The direction of the E-field is radially outward both inside and outside the sphere.
What is the magnitude of the E-field inside the sphere at a distance r<R from the center?
What is the magnitude of the E-field outside the sphere at a distance r<R from the center?
Transcribed Image Text:A hollow sphere of radius R has a uniform negative surface charge density -o on its surface and a positive point charge +Q at
its center. The charge Q is greater than the absolute magnitude of the total charge on the surface.
What is direction of the E-field inside and outside the sphere?
E=0 inside the sphere; radially inward outside the sphere.
Oradially outward inside the sphere; radially inward outside the sphere
radially outward insider the sphere; E = 0 outside the sphere
radially outward both inside and outside the sphere
What is the magnitude of the E-field inside the sphere at a distance r < R from the center? Give the answer in terms of the
symbols given in the question Q, o, R, and the constant co-
Einside =
Incorrect
What is the magnitude of the E-field outside the sphere at a distance r > R from the center?
Eoutside =
units
D
Expert Solution
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Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r> R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4πRªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4πRªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 1¬R²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 1¬R²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4πRªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 1¬R²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4πRªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r> R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 1¬R²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r> R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4πRªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4πRªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 1¬R²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 1¬R²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4πRªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 1¬R²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4πRªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r > R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 TR²G
4760
R²o
Solution
by Bartleby Expert
Follow-up Question
Step 3: Electric Field Outside the Sphere answer is incorrect. Can we try again?
Transcribed Image Text:O
Step 3: Electric Field Outside the Sphere
Electric Field Outside the Sphere (r> R)
Outside the sphere, the negative surface charge - on the sphere contributes
to the electric field.
Using Gauss's Law and a Gaussian surface in the shape of a sphere with
radius r (r > R) centered at the center of the sphere, we can find the electric
field due to the negative surface charge.
The charge enclosed within this Gaussian surface is the total charge on the
surface, which is −4Rªo.
Therefore, by Gauss's Law, the magnitude of the electric field outside the
sphere at a distance r > R from the center is:
Eoutside
1 1¬R²G
4760
R²o
Solution
by Bartleby Expert
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