Flux Through a Gaussian Surface

[abysmith18]

Homework Statement

A metal sphere of radius a is surrounded by a metal shell of inner radius b and outer radius R, as shown in the diagram below. The flux through a spherical Gaussian surface located between a and b is 1.20Q/εo and the flux through a spherical Gaussian surface just outside R is 0.80Q/εo.
a) What is the total charge on the inner sphere? (Express your answer as a multiple of Q. For example, if the total charge is 0.2Q, then input 0.2).

Homework Equations

flux=E4pi*r^2=Qenclosed/epsilon not

The Attempt at a Solution

I understand that I need to use the ratios of 1.2Q/Enot=Qenclosed/Enot and .8Q/Enot=Qenclosed/Enot
I'm just not sure what to do with these two things

 

[Delta2]

Sorry for this question but do you understand what the meaning of $Q_{enclosed}$ is? If yes you should be able to see that the $Q_{enclosed}$ by the Gaussian surface given by the problem is equal to the charge of the metal sphere of radius a. And using the first of ratios you wrote you ll easily conclude that $Q_{enclosed}$ is equal to ...

 

[haruspex]

epsilon not

It's "epsilon nought". I've heard video tutorials, from India I think, which pronounce it "not" instead of "nought".

 

[joyousvoyage]

Sorry for this question but do you understand what the meaning of $Q_{enclosed}$ is? If yes you should be able to see that the $Q_{enclosed}$ by the Gaussian surface given by the problem is equal to the charge of the metal sphere of radius a. And using the first of ratios you wrote you ll easily conclude that $Q_{enclosed}$ is equal to ...

equal to what?

 

[Delta2]

equal to what?

1.2Q of course!

 

[Steve4Physics]

Hi, Suppose a surface completely surrounds a charge of, say, 3C. What is the total flux passing through the surface?

Use Gauss's law to answer the question. If you can't do it, you might need to do a little reading or watch a YouTube video first.

After that you might be able to answer the original question. Or we'll be in a better position to help you.

 

[joyousvoyage]

1.2Q of course!

wouldn't it be -1.2Q? not 1.2Q?

 

[Steve4Physics]

wouldn't it be -1.2Q? not 1.2Q?

There is no minus sign in the equation:
$\Phi = \frac {q} {\epsilon_0}$ (equation 1)

A surface between a and b encloses the inner metal sphere.

You are told the flux through this surface is $\Phi =\frac {1.20Q}{\epsilon_0}$. Substitute this value into equation 1.

When you do this, what does this give for the charge enclosed by the surface: q = ?