Electric field and gaussian surface

In summary, a small copper spherical BB with a charge of +q is located within a larger hollow copper spherical shell with inner radius b and outer radius R, which has zero charge on it. The electric field within the BB for radii r<a is zero, as is the electric field inside the copper shell for radii satisfying b<r<R. According to Gauss's Law, the total flux of the electric vector through a closed Gaussian surface within the copper shell is also zero, implying that charge must only reside on the inside surface of the shell. Since the total charge within the shell is zero, the charge on the inner surface must also be zero, meaning that the same amount of charge must reside on the outer surface of the copper shell.
  • #1
PinkFlamingo
19
0
A small copper spherical BB of radius a is located at the center of a larer hollow copper spherical shell of inner radius b and outer radius R. A charge of +q is on the small BB. The hollow copper shell has zero charge on it.

a) What is the electric field within the BB (for radii r<a)?

b) What is the electric field inside the copper shell (for radii that stisfy b<r<R)

c) Draw a closed Gaussian surface within the copper of the shell. What is the total flux of the electric vector through this Gaussian surgace? This result implies that charge must lie on the inside surface of the spherical shell. What charge must reside on the inside surgface of the copper shell? Since the copper shell has a total charge of zero, what charge must reside on the outer surface of the copper shell?

Please help me!

So far all I've been able to figure out is that the answer to b is 0 I think?
 
Last edited:
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  • #2
These questions require you to know two things:
(1) The electrostatic field within a conductor is zero
(2) Gauss's Law, which relates the total electric flux through a closed surface to the charge within that surface

Can you state Gauss's Law? How do you apply it?
 
  • #3
Would the answer to a AND b be 0 because they're both within the conductor?
 
  • #4
PinkFlamingo said:
Would the answer to a AND b be 0 because they're both within the conductor?
Yes. The field within both conductors is zero.
 
  • #5
Would the answer to c be 0 as well, since the total charge is 0, the flux must be 0 too right?
 
  • #6
PinkFlamingo said:
Would the answer to c be 0 as well, since the total charge is 0, the flux must be 0 too right?
See my comments:
c) Draw a closed Gaussian surface within the copper of the shell. What is the total flux of the electric vector through this Gaussian surgace?
Since it's within a conductor, the field and flux must be zero.
This result implies that charge must lie on the inside surface of the spherical shell. What charge must reside on the inside surgface of the copper shell?
So what charge must lie on the inside surface? Remember: Total charge within the Gaussian surface must be zero.
Since the copper shell has a total charge of zero, what charge must reside on the outer surface of the copper shell?
The charge on the inner surface plus the charge on the outer surface must add to zero. Figure it out.
 

Related to Electric field and gaussian surface

1. What is an electric field?

An electric field is a physical quantity that describes the influence a charged object has on other charged objects in its vicinity. It is a vector quantity, meaning it has both magnitude and direction, and is measured in Newtons per Coulomb (N/C).

2. What is a gaussian surface?

A gaussian surface is a hypothetical surface that is used to calculate the electric field at a point in space. It is a closed surface that encloses a charge or charges, and is chosen based on the symmetry of the charge distribution to simplify the calculation of the electric field.

3. How is the electric field related to the charge enclosed by a gaussian surface?

According to Gauss's Law, the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. This means that the electric field at a point can be found by dividing the charge enclosed by a gaussian surface by the surface area of the gaussian surface.

4. Can a gaussian surface be any shape?

No, a gaussian surface must have a symmetrical shape to properly apply Gauss's Law and simplify the calculation of the electric field. Common shapes used for gaussian surfaces include spheres, cylinders, and cubes.

5. How is the electric field calculated for a non-uniformly charged gaussian surface?

To calculate the electric field for a non-uniformly charged gaussian surface, the surface is divided into small, uniform charge elements. The electric field is then calculated for each charge element using Coulomb's Law, and the contributions from all the charge elements are added together to find the total electric field at the point of interest.

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