Nonconducting spherical shell with uniform charge

[ooohffff]

Homework Statement

Suppose the nonconducting sphere of Example 22-4 has a spherical cavity of radius r1 centered at the sphere's center (see the figure). Assuming the charge Q is distributed uniformly in the "shell" (between r = r1 and r = r0), determine the electric field as a function of r for the following conditions.

r1 < r < r0

Homework Equations

Gauss's law

The Attempt at a Solution

I know the electric field is 0 in the inner cavity. The electric field outside of the shell is Q/(4pi*E0*r^2). How would I derive the equation for the electric field inside the shell? Charge per unit volume should be p=Q/(4/3*pi(r0-r1)^3) right?

 

[SammyS]

Homework Statement

Suppose the nonconducting sphere of Example 22-4 has a spherical cavity of radius r1 centered at the sphere's center (see the figure). Assuming the charge Q is distributed uniformly in the "shell" (between r = r1 and r = r0), determine the electric field as a function of r for the following conditions.

r1 < r < r0

Homework Equations

Gauss's law

The Attempt at a Solution

I know the electric field is 0 in the inner cavity. The electric field outside of the shell is Q/(4pi*E0*r^2). How would I derive the equation for the electric field inside the shell? Charge per unit volume should be p=Q/(4/3*pi(r0-r1)^3) right?

That charge density is incorrect.

A³ - B³ ≠ (A - B)³ .

 

[ooohffff]

That charge density is incorrect.

A³ - B³ ≠ (A - B)³ .

Mmm yes, your'e right it should be Q/(4pi/3(r0^3-r1^3)). But where do you go from there?

 

[SammyS]

Mmm yes, your'e right it should be Q/(4pi/3(r0^3-r1^3)). But where do you go from there?

How much charge there interior to a sphere of radius, r, where r₁ < r < r₀ .

 

[ooohffff]

How much charge there interior to a sphere of radius, r, where r₁ < r < r₀ .

Could you rephrase that and be more specific?

 

[ooohffff]

Would it be the same as the electric field outside the sphere?

 

[SammyS]

Would it be the same as the electric field outside the sphere?

No, electric charge is not equal to electric field.

You want to find what the electric field is interior to the "shell". That's where r₁ < r < r₀ .

You mention Gauss's Law.

 

[ooohffff]

No, electric charge is not equal to electric field.

You want to find what the electric field is interior to the "shell". That's where r₁ < r < r₀ .

You mention Gauss's Law.

Right its

∫E⋅da=q_enc/ϵ₀

So I keep getting E4πr^2=q_enc/ϵ₀

so that would mean E would equal the same as the electric field as the outside of the sphere, which I know is wrong. What am I doing wrong?

 

[SammyS]

Right its

∫E⋅da=q_enc/ϵ₀

So I keep getting E4πr^2=q_enc/ϵ₀

so that would mean E would equal the same as the electric field as the outside of the sphere, which I know is wrong. What am I doing wrong?

For r between r₁ and r₀, what do use for your Gaussian surface?

 

[ooohffff]

For r between r₁ and r₀, what do use for your Gaussian surface?

Would it be r1^2 instead of r^2?

 

[SammyS]

Would it be r1^2 instead of r^2?

No.

If you want to know the electric field at a distance r from the origin, use s sphere of radius r.

How much electric charge is enclosed within this sphere?

In the case of your problem the only charge inside this sphere resides in the spherical shell with inner radius r₁ and outer radius r. Right ?

So, how much charge is that ?

 

[ooohffff]

So if charge density =

Q/ (4π/3 ( r0³-r1³)) = q_enc/(4π/3 (r³-r1³)).

Then,

Q(r³-r1³) = q_enc (r0³-r1³)

q_enc = [Q(r³-r1³)] / (r0³-r1³)

E4πr² = Q(r³-r1³)] / ε₀(r0³-r1³)

E = Q(r³-r1³)] / 4πr²ε₀(r0³-r1³)

 

[ooohffff]

Can someone verify this?

 

[SammyS]

Can someone verify this?

Patience.

So if charge density =

Q/ (4π/3 ( r0³-r1³)) = q_enc/(4π/3 (r³-r1³)).

Then,

Q(r³-r1³) = q_enc (r0³-r1³)

q_enc = [Q(r³-r1³)] / (r0³-r1³)

E4πr² = Q(r³-r1³)] / ε₀(r0³-r1³)

E = Q(r³-r₁³)] / [4πr²ε₀(r₀³-r₁³) ]

That's the right idea. To be perfectly correct, the entire denominator should be enclosed in parentheses, else use some other grouping symbol.

 

[ooohffff]

Patience.
That's the right idea. To be perfectly correct, the entire denominator should be enclosed in parentheses, else use some other grouping symbol.

Haha, thank you SammyS!