Electric field of a sphere and a shell

[kirito]

Homework Statement

A sphere with radius r is uniformly charged with a charge q .
The sphere is coated with a spherical shell with thickness 2r so between r and 3r .
The shell is uniformly charged with a charge -q

What is the electric field at a distance 2r from the center of the sphere?

Relevant Equations

I used gauss law but i am not sure at 2a is the answer 0 or $$ kq/(4r^2) $$in the radial direction
What is considered the enclosed q in this case an explanation is appreciated

q encolsed =0
Second case q enclosed q by gauss law
At 2r

 

[kuruman]

I used gauss law but i am not sure at 2a is the answer 0 or kq/(4a^2) in the radial direction

What is a? Did you mean r?

What is considered the enclosed q in this case an explanation is appreciated

Enclosed by what? What is your Gaussian surface?

Please show you work. Start by saying "Consider a Gaussian sphere of radius = (enter radius of sphere) . . . "

 

[kirito]

What is a? Did you mean r?

Enclosed by what? What is your Gaussian surface?

Please show you work. Start by saying "Consider a Gaussian sphere of radius = (enter radius of sphere) . . . "

So i used a spherical gaussian sphere of radius 2r to coat the small sphere of radius r inside

$$\oint \mathbf{E} \cdot d\mathbf{A} = E \cdot 4\pi (2r)^2 = \frac{q}{\epsilon_0}
$$


$$E \cdot 4\pi (2r^2) = \frac{q}{\epsilon_0}


E = \frac{q}{16\pi \epsilon_0 r^2}$$

 

[jbriggs444]

How much charge is enclosed by the surface of radius $2r$.

 

[kirito]

How much charge is enclosed by the surface of radius $2r$.

From what i deduced just q

 

[kuruman]

$$E \cdot 4\pi (4r^2) = \frac{q}{\epsilon_0}E = \frac{q}{16\pi \epsilon_0 r^2}$$

This equation says that
$$E \cdot 4\pi (4r^2) = \frac{q}{\epsilon_0}E$$Check your algebra. Then draw a good diagram with the Gaussian surface at $2r$ and find the total charge enclosed.

 

[kuruman]

From what i deduced just q

Please show how you deduced that.

 

[kirito]

Please show how you deduced that.

i seem to be bad at phrasing and latex as well i hope this makes my approach clear,
The gaussian sphere i used has a radius of 2r from the center seems to enclose just the small sphere while the -q is on the shell so i assumed when substituting in the equation it is not enclosed
$$\oint Eda= E4\pi(2r)^2=q enclosed /\epsilon$$
I moved everything to the right and let E on one side and thus got whats supposed too bee the electric field
Then I started doubting this approach and thinking maybe -q is actually enclosed too by the spherical shape i used

 

[kuruman]

Then I started doubting this approach and thinking maybe -q is actually enclosed too by the spherical shape i used

1. Positive charge is enclosed from 0 to r
2. Negative charge is enclosed from r to 3r
You need to find how much negative charge is enclosed from r to 2r and add it to the positive charge to find the net.

 

[kuruman]

Here is a drawing. Positive charge is red of radius r, negative charge is green of radius 3r. The Gaussian surface at 2r is represented by the dashed white line.

 

[kirito]

I see i should really have drawn a better drawing i totally forgot about -q being from r to 2r thank you so much i will try again

 

[kuruman]

. . . i totally forgot about -q being from r to 2r . . .

No. Read carefully what you wrote below.

The sphere is coated with a spherical shell with thickness 2r so between r and 3r
The shell is uniformly charged with a charge -q

This says that the entire green volume in post #10 contains charge -q.

 

[kirito]

No. Read carefully what you wrote below.

This says that the entire green volume in post #10 contains charge -q.

True i was fixated on the drawing i drew to simplify the question and since -q was written near the line of the gaussian shape i considered it from 2r to 3r and started debating if it is inside the spherical shape that has 2r as radius or is it on the outer part ,
I think i will try to draw a gaussing shape that has radius r and fill it with -q and a gaussian shape with 2r and -q and substract the smaller one from the bigger one so that the amount of q i have is just between r and 2r , will update after taking a small break

 

[kirito]

so here what I have done
$$ E 4\pi(2r)^2=\frac{\frac{-q}{(4/3 \pi (3r)^3 -4/3 \pi r^3)}*(4/3 \pi (2r)^3-4/3 \pi r^3)+\frac{q}{4/3 \pi(r)^3} 4/3\pi (r)^3 }{\epsilon}$$

 

[kirito]

so $$ E =\frac{19q}{\epsilon26 *16 r^2 \pi}$$

 

[jbriggs444]

so here what I have done
$$ E 4\pi(2r)^2=\frac{\frac{-q}{(4/3 \pi (3r)^3 -4/3 \pi r^3)}*(4/3 \pi (2r)^3-4/3 \pi r^3)+\frac{q}{4/3 \pi(r)^3} 4/3\pi (r)^3 }{\epsilon}$$

That is a pretty verbose formula. The underlying concept is better presented in words.

You are not concerned with the exact volume for the spherical shell between $r=1$ and $r=2$. You are only concerned with what fraction that volume makes relative to the total volume from $r=1$ to $r=3$.

That is, you only care about proportionality. All those $\frac{4}{3}\pi r^3$ multipliers are wasted $\LaTeX$.

In terms of proportionality, the volumes of concentric spheres of radius 1, 2 or 3 go as 1, 8 and 27.

The incremental volumes of the three spherical shells go as 1, 7 and 19.

The shell from radius 1 to radius 3 goes as 26 with total charge $-q$.
The shell from radius 1 to radius 2 goes as 7 with charge $\frac{-7}{26}q$
The total charge including the $+q$ charge from radius 0 to 1 is then $\frac{19}{26}q$

Which matches the figure embedded in your result.

 

[kirito]

That is a pretty verbose formula. The underlying concept is better presented in words.

You are not concerned with the exact volume for the spherical shell between $r=1$ and $r=2$. You are only concerned with what fraction that volume makes relative to the total volume from $r=1$ to $r=3$.

That is, you only care about proportionality. All those $\frac{4}{3}\pi r^3$ multipliers are wasted $\LaTeX$.

In terms of proportionality, the volumes of concentric spheres of radius 1, 2 or 3 go as 1, 8 and 27.

The incremental volumes of the three spherical shells go as 1, 7 and 19.

The shell from radius 1 to radius 3 goes as 26 with total charge $-q$.
The shell from radius 1 to radius 2 goes as 7 with total charge $\frac{-7}{26}q$
The total charge including the $+q$ charge from radius 0 to 1 is then $\frac{19}{26}q$

thank you a lot ,this is really clear and makes the solution apparent as for the $\frac{4}{3}\pi r^3$ being wasted latex I agree , I just feared that writing without them would make it harder to understand what's the equation is about from finding the density of the electrons and then the total electrons in the volume unit of interest so that if I did a mistake it would be easier to pinpoint , next time I will try to explain an equation using words