How do electric flux and radius affect each other in a Gaussian sphere?

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The discussion focuses on deriving expressions for electric flux through a Gaussian sphere in relation to its radius. For a radius r less than a (the radius of the insulating sphere), the electric flux is calculated using the charge density and volume, while for r greater than a, it involves the total charge Q. Gauss's law is applied to determine the total charge enclosed within the Gaussian surface. The problem emphasizes the relationship between electric flux and the radius of the Gaussian surface. Understanding these concepts is crucial for solving related physics problems effectively.
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Homework Statement


I have a new question if anyone could help...

The question states:
Find an expression for the electric flux passing through the surface of the Gaussian sphere as a function of r for r < a. (Use epsilon_0 for 0, rho for , Q, a, and r as necessary.)

Find an expression for the electric flux for r > a. (Use epsilon_0 for 0, p for , Q, a, and r as necessary.)

a, being the radius of the sphere and r being the given radius that is either <a, or >a
 
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Present the problem exactly as given. Don't leave anything out.
 
sorry, here is the full problem...

An insulating solid sphere of radius a has a uniform volume charge density and carries a total positive charge Q. A spherical Gaussian surface of radius r, which shares a common center with the insulating sphere, is inflated starting from r = 0.

(a) Find an expression for the electric flux passing through the surface of the Gaussian sphere as a function of r for r < a. (Use epsilon_0 for 0, rho for , Q, a, and r as necessary.)

(b) Find an expression for the electric flux for r > a. (Use epsilon_0 for 0, p for , Q, a, and r as necessary.)
 
Apply Gauss's law. What you'll need to do is find the total charge contained within the Gaussian surface as a function of r. That's just charge density times volume. (Figure out the charge density from the given information.)
 
thanks. i did that and it worked out!
 
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