Electric flux changing forumla variables

AI Thread Summary
A point charge Q at the center of a sphere produces a specific electric flux. Doubling the radius of the sphere does not change the electric flux, as it is determined solely by the total charge within the boundary, according to Gauss' Law. The formula for electric flux indicates that the radius does not affect the total flux since it cancels out when calculating. The correct understanding is that the electric flux remains constant regardless of the sphere's radius. Thus, the electric flux remains equal to Q/ε₀, confirming that the initial answer was incorrect.
kirby2
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A point charge Q at the center of a sphere of radius R produces an electric flux of (flux symbol) coming out of the sphere. If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be?

LINK TO ORIGINAL PROBLEM: http://i.imgur.com/e8Tmt.png

ATTEMPT: using the formula: flux=E(4(pi)r^2) I got (flux symbol)/4 as my answer. Is this right?
 
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kirby2 said:
A point charge Q at the center of a sphere of radius R produces an electric flux of (flux symbol) coming out of the sphere. If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be?

LINK TO ORIGINAL PROBLEM: http://i.imgur.com/e8Tmt.png

ATTEMPT: using the formula: flux=E(4(pi)r^2) I got (flux symbol)/4 as my answer. Is this right?

Have you studied Gauss' Law? What does it say about the flux and the total charge contained within a boundary?
 
well flux is = EA where A = 4(pi)r^2 and E = (kq)/r^2. so, it looks like the r^2 will cancel when multiplied, meaning that changing the radius doesn't change the flux. is this sound logic?
 
kirby2 said:
well flux is = EA where A = 4(pi)r^2 and E = (kq)/r^2. so, it looks like the r^2 will cancel when multiplied, meaning that changing the radius doesn't change the flux. is this sound logic?

That'll do :smile:

FYI, Gauss' Law states that the total flux through a closed bounding surface is entirely determined by the total charge contained within that surface:

\Phi = \frac{Q}{\epsilon_0}

That's probably a more direct route to the same answer.
 
thank you very much
 
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