Understanding Gauss's Law and Electric Field Distributions

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In summary, In answer 2, adding a point charge at the origin increases the field strength at point P, while adding a point charge at point (2R, 0) decreases the field strength at point P. In answer 3, adding a uniformly charged nonconducting sphere with charge -Q and radius 1/2R has the same effect as adding a point charge at the origin.
  • #1
PhysicsNewb
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Hi, I had three problems to do that are multiple choice, here they are and my answer with them. Please tell me if some are wrong.

1. Spherical conductor of radius R carries char +Q, which statement(s) are correct?

I. Electric field outside the surface of the conductor is the same as the field outside a uniformly charged nonconducting sphere of same radius and charge

II. Electric field inside the surface of the conductor is the same as the field inside a uniformly charged nonconducting sphere of same radius and charge

III. Nonspherical Gaussian surface drawn within the conductor will have a ney electric flux through it

I said that only statement I. was correct.


2. Gauss's law could easily be used to solve for the electric field outside all of the following charge distributions except

A.) charged nonconducting sphere
B.) long straight line of charge
C.) long charged conducting cylinder
D.) charged nonconducting cube
E.) charged nonconducting plane

I chose B. was correct.

3. Uniformly charged nonconducting sphere of radius R, charge +Q, is centered at the origin of a cartesian plane. Field strength at Point P with coordinates (R,0) just outside the surface will double if:

I. Add a point charge +Q at the origin
II. Add a point charge -Q at point (2R, 0)
III. Add a uniformly charged nonconducting sphere with charge -Q and radius 1/2R with its center at the point (2R, O)

I chose I. only as the answer.

Are these right? If not, please help..
 
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  • #2
PhysicsNewb said:
1. Spherical conductor of radius R carries char +Q, which statement(s) are correct?

I. Electric field outside the surface of the conductor is the same as the field outside a uniformly charged nonconducting sphere of same radius and charge

II. Electric field inside the surface of the conductor is the same as the field inside a uniformly charged nonconducting sphere of same radius and charge

III. Nonspherical Gaussian surface drawn within the conductor will have a ney electric flux through it

I said that only statement I. was correct.
Right.

2. Gauss's law could easily be used to solve for the electric field outside all of the following charge distributions except

A.) charged nonconducting sphere
B.) long straight line of charge
C.) long charged conducting cylinder
D.) charged nonconducting cube
E.) charged nonconducting plane

I chose B. was correct.
Incorrect. Hint: Which charge distribution lacks a simple symmetry?

3. Uniformly charged nonconducting sphere of radius R, charge +Q, is centered at the origin of a cartesian plane. Field strength at Point P with coordinates (R,0) just outside the surface will double if:

I. Add a point charge +Q at the origin
II. Add a point charge -Q at point (2R, 0)
III. Add a uniformly charged nonconducting sphere with charge -Q and radius 1/2R with its center at the point (2R, O)

I chose I. only as the answer.
Incorrect. Add the contributions from each charge to find the new field at point P.
 
  • #3
I'm going to revise my answers to make number 2. answer D and number 3 answer's I. and II. I think i worked it right this time.
 
Last edited:
  • #4
PhysicsNewb said:
I'm going to revise my answers to make number 2. answer D and number 3 answer's I. and II.
For question 3, what's the difference between answers II and III (as far as the field at point P is concerned)?
 

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