Gauss's Law and a conducting sphere

[Alem2000]

"A solid conducting sphere with radius $$R$$ carries a posative total charge $$Q$$. The sphere is surrounded by an insulating shell with inner radius $$R$$ and outer radius $$2R$$. The insulating shell has a uniform charge density $$\rho$$ a) find the value of $$\rho$$ so that the net charge of the entire system is zero b) if $$\rho$$ has the value found in part (a), fnd the electric field (magnitude and direction) in each of the regions
$$0<r<R$$ $$R<r<2R$$ and $$r>2R$$"

Calculating charge in terms of $$\rho$$ i got
$$\sum Q=\frac{-28\pi\rho R^3}{3}$$

now my problem is trying to fine the $$\vec{E}$$ below is my work
$$\oint\vec{E}d\vec{A}=\frac{Q_inc}{\epsilon_0}$$
there is an electric field only between
$$R<r<2R$$
$$\vec{E}=\frac{Q}{4\pi R^2 \epsilon_0}$$
and after solving my above value for
$$\rho$$ in terms of $$Q$$
I got

$$\vec{E}=\frac{7R\rho}{3}$$

which is soo wrong, I am sure I did some of this problem correctly..the part I don't understand is how I would find the electric field? Can anyone please help?

 

[Dr Transport]

You had the charge density in the shell correct but forgot that

$$Q_{inc} = Q - 4\pi \int_{R} ^{R'} \rho r^{2} dr$$

not just the charge of the spherical shell...

 

[thepatientmental]

First, let's address part (a) of the problem. We are given a solid conducting sphere with a positive charge Q and an insulating shell with a uniform charge density \rho surrounding it. In order for the net charge of the entire system to be zero, the total charge of the insulating shell must be equal and opposite to the total charge of the conducting sphere. This means that we can set up the following equation:

Q + Q_shell = 0

Solving for Q_shell, we get:

Q_shell = -Q

Now, in terms of \rho, we can express the charge of the insulating shell as:

Q_shell = \rho * volume of shell

Since the insulating shell has an inner radius of R and an outer radius of 2R, its volume can be calculated as:

volume of shell = \frac{4}{3}\pi(2R)^3 - \frac{4}{3}\pi R^3 = \frac{28}{3}\pi R^3

Plugging this into our equation for Q_shell, we get:

-Q = \rho * \frac{28}{3}\pi R^3

Solving for \rho, we get:

\rho = \frac{-3Q}{28\pi R^3}

Now, for part (b) of the problem, we need to find the electric field in each of the given regions. Let's start with the region 0<r<R, which is inside the conducting sphere. Since this is a solid conducting sphere, the electric field inside it will be zero. This is because the charges in the conducting sphere will redistribute themselves to cancel out any external electric field.

Next, for the region R<r<2R, which is inside the insulating shell, we can use Gauss's Law to find the electric field. The Gaussian surface we will use is a spherical shell with radius r. The electric field will be constant on this surface, since the charge density \rho is uniform. The flux through this surface will be:

\Phi = \vec{E} * 4\pi r^2

And according to Gauss's Law, this should be equal to the total enclosed charge divided by \epsilon_0. The total enclosed charge in this case is the charge of the insulating shell, which we found to be -Q. So we have:

\Phi = \vec{E} * 4\pi r^