# Gauss' Law Question

How would you solve this:

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!!!

## Answers and Replies

siddharth
Homework Helper
Gold Member
You wrote the title as "Gauss Law Question". Do you know what Gauss' law says? Can you apply it here?

I know it is Gauss' Law because that is what the website I got it from said. I have looked everywhere and applied the Gauss' Law how I thought I should apply it but I keep getting the wrong answer and have no idea how they got that answer. So that is why I need someone to explain it for me if they know.

Hootenanny
Staff Emeritus
Science Advisor
Gold Member
delongk said:
I know it is Gauss' Law because that is what the website I got it from said. I have looked everywhere and applied the Gauss' Law how I thought I should apply it but I keep getting the wrong answer and have no idea how they got that answer. So that is why I need someone to explain it for me if they know.
Perhaps if you showed your working we could point out where you have gone wrong...

Well, I am confused as to what Gauss Equation to use... I have tried q=Eo*flux and then I tried flux=q/4pi*k and then I tried flux=k*q/r^2... none of them turn out correct.

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

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.

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

Last edited:
siddharth
Homework Helper
Gold Member
443 C, is an awful lot of charge. Are you sure you got the numbers right from the question?

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

I copied and pasted it so I know that is right. And I keep getting the wrong answer for ALL of the examples... which are all similar to that one.

Another example on the webpage is:

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.

siddharth
Homework Helper
Gold Member
Ok, Gauss' law is

$$\int \vec{E}.d\vec{a} = \frac{q_{enc}}{\epsilon_0}$$

That's the equation you're going to use for these problems.
Now, note that
(i) The value of $\int \vec{E}.d\vec{a}$ is the flux passing through the surface.
(ii) Using Gauss' law, it follows from the previous step that $\frac{q_{enc}}{\epsilon_0}$ 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 $\int \vec{E}.d\vec{a}$ can be simply reduced to E times A. Can you figure out why, and solve the problems from here?

no i have no idea what to do... i need step by step. and everything i read said you dont need the radius.

Mute
Homework Helper
I think the problem is that whoever calculated the answers on that site is incorrectly using the Coloumb force constant {1/(4*Pi*epsilon_0)} instead of the permittivity of free space (epsilon_0) in their calculations. For instance, they ask this question:

"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.