Solving a Charge at the Center of a Sphere Problem

In summary, Gauss' law can be used to determine the electric field at every point in space for a hollow electrically neutral conducting sphere with a positive charge at the center. The two radii given (R1 and R2) mark off three distinct regions and must be taken into account when constructing a Gaussian surface. The electric field can be expressed as a function of r (the distance from the center of the sphere), with different expressions for r < R1, R1 < r < R2, and r > R2.
  • #1
BOAS
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Homework Statement



A positive charge q is placed at the center of a hollow electrically neutral conducting sphere (inner radius R1 9cm, outer radius R2 10 cm.

Using Gauss' law determine the electric field of every point in space, as a function of r (the distance from the center of the sphere). Only the algebraic expression is required.

Homework Equations

The Attempt at a Solution



I'm unsure of how to go about this problem, all the examples I have seen discuss thin spheres.

A positive charge at the center of the sphere would induce a positive charge on the surface of the sphere, whilst a negative one would be induced on the inner surface.

I think it is safe to assume the charge at the center is a point and therefore the charge will be evenly distributed creating a symmetrical electric field.

[itex]\Phi _{0} = ( \Sigma \cos \phi) \Delta A[/itex]

I know I need to 'construct' a gaussian surface with radius r (r>R2) concentric with the shell, but I don't know how to use the information about the two radii I was given - Are they important here?

Since the electric field is everywhere perpendicular to the gaussian surface, [itex]\phi = 0^{o}[/itex] and [itex] \cos \phi = 1[/itex].

The electric charge has the same value all over the surface, so we can say that;

[itex]\Phi _{0} = E( \Sigma \Delta A) = E(4 \pi r^{2}) [/itex]

Setting [itex]\Phi _{0} = \frac{q}{\epsilon _{0}} [/itex]

We can say that [itex] E = \frac{q}{4 \pi \epsilon _{0} r^{2}}[/itex]

For r > R2 since Gauss' law also shows us that there is no net charge inside the sphere.

The above makes sense to me, but at no point did I use the radii given to me in the question...

What am I doing wrong?

Thanks!

BOAS
 
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  • #2
BOAS said:
I know I need to 'construct' a gaussian surface with radius r (r>R2) concentric with the shell, but I don't know how to use the information about the two radii I was given - Are they important here?
They are important because they mark off three distinct regions: r < R1; R1 < r < R2; r > R2.

You only dealt with the last region. What about the others?
 
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  • #3
Ok, that was surprisingly obvious.

I'll have three distinct expressions, does that satisfy the question of finding the electric field of every point in space?

I suppose that it does, but on first reading of the question I was expecting a single expression.

Thanks for your help, I'm confident I can do this now.
 

FAQ: Solving a Charge at the Center of a Sphere Problem

1. How do you determine the electric field at the center of a charged sphere?

To calculate the electric field at the center of a charged sphere, you can use the formula E = kQ/R^2, where k is the Coulomb's constant, Q is the charge of the sphere, and R is the radius of the sphere. This formula is derived from Coulomb's law, which states that the electric field is directly proportional to the charge and inversely proportional to the distance squared.

2. Can the electric field at the center of a sphere be zero?

Yes, the electric field at the center of a sphere can be zero if the sphere has an equal amount of positive and negative charge, also known as a neutral sphere. In this case, the electric fields from the positive and negative charges cancel each other out at the center.

3. How does the electric field change if the charge is moved from the center to the surface of the sphere?

If the charge is moved from the center to the surface of the sphere, the electric field at the center will decrease. This is because the distance between the charge and the center increases, according to the formula E = kQ/R^2. However, the electric field at the surface of the sphere will increase.

4. Can the electric field at the center of a sphere be negative?

Yes, the electric field at the center of a sphere can be negative if the charge of the sphere is negative. This indicates that the direction of the electric field is towards the center, which is opposite to the direction of a positive charge.

5. How does the electric field at the center of a sphere change with the change in charge or radius?

The electric field at the center of a sphere is directly proportional to the charge and inversely proportional to the square of the radius. This means that if the charge of the sphere increases, the electric field at the center will also increase. Similarly, if the radius of the sphere increases, the electric field at the center will decrease.

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