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A small charge of 443 C is at the center of a 7.97 cm radius ball. How much flux passes through the ball's surface?

The answer is 4.922 E-8 N.m2/C

I don't know how to get this answer. Please explain. Thank you!!!

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- Thread starter delongk
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- #1

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A small charge of 443 C is at the center of a 7.97 cm radius ball. How much flux passes through the ball's surface?

The answer is 4.922 E-8 N.m2/C

I don't know how to get this answer. Please explain. Thank you!!!

- #2

siddharth

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- #4

Hootenanny

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Perhaps if you showed your working we could point out where you have gone wrong...delongk said:

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What is the equation to use for a point charge in a sphere?

- #6

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I keep finding that flux= q/Eo and so I put 443 in for q and 8.85*10^-12 for Eo and get 5.006*10^13.

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Assuming that you are familiar with calculus, you can have a look at http://en.wikipedia.org/wiki/Gauss_law. The flux is indeed [itex]q/\epsilon_{0}[/itex] in your case Do you have a doubt about the flux being independent of [itex]R[/itex]? If so, the wiki article/a reading from any standard textbook should clear it.

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- #8

siddharth

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You calculated the flux as q_enc/Eo, and that should work.

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- #10

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A small charge is at the center of a 9.77 cm radius ball. If 8.322 E-8 N.m2/C passes through the ball's surface, how much charge is at the center?

and the answer is 749C.

I don't get this one either. I am assuming the use the same equation.

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http://www.dctech.com/physics/help/problems.php?problem=electric-gauss

- #12

siddharth

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[tex] \int \vec{E}.d\vec{a} = \frac{q_{enc}}{\epsilon_0} [/tex]

That's the equation you're going to use for these problems.

Now, note that

(i) The value of [itex]\int \vec{E}.d\vec{a}[/itex] is the flux passing through the surface.

(ii) Using Gauss' law, it follows from the previous step that [itex]\frac{q_{enc}}{\epsilon_0}[/itex] is also the flux passing through the surface.

So, you've got 2 ways to calculate the flux.

Why have they given the radius? Because you can calculate the area of the ball. Notice that, when a charge is at the center of a sphere, the integral [itex]\int \vec{E}.d\vec{a}[/itex] can be simply reduced to E times A. Can you figure out why, and solve the problems from here?

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- #14

Mute

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"A small charge of 45 C is at the center of a ball. If 5 E-9 N.m2/C passes through the ball's surface, what is the radius of the ball (if it matters)? "

Now, obviously for that question the answer is that the radius doesn't matter, but if you use their values for charge and the flux, and plug into Flux = charge/k, solving for k you get 9x10^9, which is 1/(4*Pi*epsilon_0), not epsilon_0.

Also, if in the original question you asked, if you say Flux = 443/(9x10^9), you'll find you get their answer.

So, you were doing everything correctly; it's the website that is wrong.

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