Find the charge and electric field.

[sonutulsiani]

Homework Statement

A sphere of radius R has volume charge density = B/r for r < R, where B is a constant and = 0 for r > R. (a) Find the total charge on the sphere. (b) Find the expressions for the electric field inside and outside the charge distribution. (c) Sketch the magnitude of the electric field as a function of the distance r from the sphere’s center

Homework Equations

The Attempt at a Solution

I got the charge by integrating (4 pi B r dr) over 0 to R as (2 pi B [R^2]). For the part b, that's where I am confused.The solution that I have and the answer that I got are different as the approach is different.

What I have done is:

Q/(Q inside) = (4/3 pi R^3)/(4/3 pi r^3). Substituting Q in this equation from above and using the Gauss' law, I got E = (B r)/(2 R ε0)


But in the solution:

For r<R, they have just simply substituted Qinside = (2 pi B [r^2]) and got E = B/(2 ε0)
For r>R, they have just simply substituted Qinside = (2 pi B [R^2]) and got E = [B(R^2)]/[2 (r^2) ε0]

What here I am confused is 2 things.

1. How does the total charge become the inside charge as the solution says.
2. Notice the Q inside for r<R, it has small r but r>R has big R.

 

[gabbagabbahey]

What I have done is:

Q/(Q inside) = (4/3 pi R^3)/(4/3 pi r^3). Substituting Q in this equation from above and using the Gauss' law, I got E = (B r)/(2 R ε0)

Whoa, slow down there big fella... What exactly is "Q inside" supposed to represent? What are you using for your Gaussian surface? How are you substituting "Q/(Q inside)" into Gauss' law when Gauss' law involves only the charge enclosed by your Gaussian surface and not this obscure ratio "Q/(Q inside)"?

If the charge inside a sphere of radius $R$, with charge density $\rho(r)=B/r$ is given by

$$Q=4\pi\int_0^RBrdr=2\pi B R^2$$

then wouldn't the charge enclosed in a concentric spherical section of that sphere, of radius $r<R$ be given by

$$Q=4\pi\int_0^r Br'dr'=2\pi B r^2$$

?

 

[sonutulsiani]

In simple words, if we use a sphere as Gaussian surface, then to find for r<R, we have to enclose the Gaussian surface inside the original sphere. That's what I have been thinking all the time. Isn't it right?
And then to find the Qinside of the Gaussian surface, I used the ratio of Q and volume of the 2 spheres.

 

[sonutulsiani]

In simple words, there will be a spherical gaussian surface which will be enclosed within the original sphere. That's what I have been thinking all the while, isn't it correct?

And to find the Q inside of the gaussian surface, I took the ratio of Q and volume of the original sphere and the gaussian sphere.

Please explain me if this is wrong.

 

[gabbagabbahey]

In simple words, if we use a sphere as Gaussian surface, then to find for r<R, we have to enclose the Gaussian surface inside the original sphere. That's what I have been thinking all the time. Isn't it right?

Right.

And then to find the Qinside of the Gaussian surface, I used the ratio of Q and volume of the 2 spheres.

No, this would be true if the charge was uniformly distributed throughout the sphere's volume, but it isn't.

You need to calculate $Q_{inside}$ the same way you calculated $Q$; by integrating the given charge density over the volume enclosed by your Gaussian surface.

 

[sonutulsiani]

Ohhhhh ok!

So when r<R, the gaussian sphere will be inside the main sphere and the charge taken into consideration will only be that of the gaussian sphere as it is inside and it will be (2 pi B r^2)

And when r<R, the gaussian sphere outside and the charge taken into consideration will be that of the original sphere which will be (2 pi B R^2)

Correct? You made a huge problem really easy for me by saying this :

Right.



No, this would be true if the charge was uniformly distributed throughout the sphere's volume, but it isn't.

 

[gabbagabbahey]

Ohhhhh ok!

So when r<R, the gaussian sphere will be inside the main sphere and the charge taken into consideration will only be that of the gaussian sphere as it is inside and it will be (2 pi B r^2)

And when r<R, the gaussian sphere outside and the charge taken into consideration will be that of the original sphere which will be (2 pi B R^2)

Correct?

Assuming that the stuff in red is a typo, then yes, that's correct.

 

[sonutulsiani]

Oh lol
Yes my bad.. Thanks a lot!

 

[sonutulsiani]

Hey tell me one thing.
To find r<R we integrated it from 0 to r. How do you know that?

I mean E inside a sphere is sometimes 0 also, when there is no charge inside a sphere.

In this question it doesn't say if the charge is uniformly distributed or if it's inside the sphere. It only says the charge on sphere, right? So shouldn't E=0 when r<R?

 

[gabbagabbahey]

You are told that there is a non-zero volume charge density for r<R, that means there is charge there.