Inner/Outer Spherical Shells - Gauss' Law

In summary, 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.
  • #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?
 
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  • #2
can you please explain why did you choose the last three statements as T,F,T ..
 
  • #3
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
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
Thanks! That was right!
 

1. What is Gauss' Law?

Gauss' Law is a fundamental law in physics that relates the electric field at a point to the charge enclosed by a closed surface surrounding that point. It is named after the German mathematician and physicist Carl Friedrich Gauss.

2. How does Gauss' Law apply to inner and outer spherical shells?

Gauss' Law can be applied to both inner and outer spherical shells. For an inner spherical shell, the electric field at any point inside the shell is zero, while for an outer spherical shell, the electric field at any point outside the shell is the same as if all the charge of the shell were concentrated at its center.

3. What is the difference between an inner and outer spherical shell?

An inner spherical shell is a hollow sphere with a charged surface that encloses a space within it, while an outer spherical shell is a hollow sphere with a charged surface that surrounds a space outside of it. In an inner spherical shell, the electric field is zero inside the shell, while in an outer spherical shell, the electric field is the same as if all the charge were concentrated at the center of the shell.

4. How is the electric field calculated using Gauss' Law for inner and outer spherical shells?

To calculate the electric field using Gauss' Law for an inner spherical shell, we use the equation E = Qenclosed/ε_0, where Qenclosed is the total charge enclosed by the shell and ε_0 is the permittivity of free space. For an outer spherical shell, we use the equation E = Q/(4πε_0r^2), where Q is the total charge of the shell and r is the distance from the center of the shell to the point where the electric field is being calculated.

5. How does the electric field change as you move from the inside to the outside of an inner spherical shell or from the outside to the inside of an outer spherical shell?

For an inner spherical shell, the electric field is zero inside the shell and increases as you move towards the surface of the shell. For an outer spherical shell, the electric field decreases as you move away from the surface of the shell and becomes zero at infinity.

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