Inner/Outer Spherical Shells - Gauss' Law

  • #1
singinglupine
15
0
A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d. The inner shell has a total charge of -1q and the outer shell has a total charge of +4q.

[URL]http://capa7.phy.ohiou.edu/res/msu/physicslib/msuphysicslib/53_Efield_Calculus_Gauss/graphics/prob11_condshells.gif[/URL]

Select True or False for the following statements.
The radial component of the electric field in the region c < r < d is given by -1q/(4πε0r2).
The total charge on the inner surface of the large shell is +1q.
The total charge on the outer surface of the small shell is zero.
The total charge on the outer surface of the large shell is +3q.
The radial component of the electric field in the region r < a is given by +3q/(4πε0r2).
The electric field in the region r > d is zero.
The total charge on the inner surface of the small shell is zero.

So I thought F,T (only influenced by the smaller inside shell, but does it go the opposite sign, or stay as a -1q charge?),F,T,T,F,T. The radial component ones I don't really understand though. Can someone point out which ones may be wrong so I can think them through a bit more?
 
Last edited by a moderator:

Answers and Replies

  • #2
thebigstar25
286
0
can you please explain why did you choose the last three statements as T,F,T ..
 
  • #3
singinglupine
15
0
For the third to last, I thought since the electrical field pointed inwards for the negatively charged ring, and outwards for the positively charge ring, the region in the total center would be influence by the outer ring pushing towards the center with +4q and the inner ring pushing outwards with -1q so that you could just subtract the two. I'm not sure about this though.
For the second to last one, outside the ring you would have the outwards directing field from the positive outer ring, and a inwards directing field from the inner negative ring, so they would cancel out in opposite directions, but since the negative value is less, I thought you would have an overall positive field outside the sphere.
For the last one, the inner surface of the small shell would not be affected by anything since it's in the inside. I thought the outer surface of the small shell would be affected by -1q, same for the inner surface of the outer shell, and the outer surface of the large shell be affected by both -1q and +4q.
 
  • #4
thebigstar25
286
0
first i would tell not to take what I am going to say as final ..

i think the explanation you provided for the last two is reasonable ..

but for "The radial component of the electric field in the region r < a is given by +3q/(4πε0r2)" , I disagree with what you said ..

im going to say what I am thinking of (again, don't take what I am saying as final) ..

for the bigger shell (which is with positive charge), the electric field should be zero inside .. so if we want to consider the electric field with r<a i think it should be something different ..
 
  • #5
singinglupine
15
0
Thanks! That was right!
 

Suggested for: Inner/Outer Spherical Shells - Gauss' Law

Replies
11
Views
711
Replies
10
Views
688
Replies
19
Views
630
  • Last Post
Replies
20
Views
821
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
445
Replies
3
Views
916
  • Last Post
Replies
1
Views
470
Top