# Help with Gauss's Law please

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1. Jun 24, 2017

### Chemmjr18

1. The problem statement, all variables and given/known data
A charge of -30 μC is distributed uniformly throughout a spherical volume of radius 10.0 cm. Determine the electric field due to this charge at a distance of (a) 2.0 cm, (b) 5.0 cm, and (c) 20.0 cm from the center of the sphere.

2. Relevant equations
Eq. (1): E⋅A=qenc
Eq. (2): qenc=q⋅(r/R)3

In qenc, r is the radius of my Gaussian surface and R is the radius of the actual sphere, 10.0 cm.

3. The attempt at a solution
(a) E= -5.8E8 N/C

(b) E= -1.35E7 N/C

(c) This is where I'm a bit stuck. If I let the radius of my Gaussian surface be 20.0 cm, then all of the actual sphere will be enclosed in my surface. Therefore, qenc would be -30 μC. However, if I use Eq. (2), I get that qenc is -2.4E-6 C which wouldn't really make any sense. Why would there be more charge than what's given? Using what I feel is the more rational option (i.e. letting qenc be -30 μC, I get the following answer:
E= -6.7E6 N/C

If I'd used Eq. (2) to find qenc, I would've gotten E= -5.4E7 N/C. This doesn't make any sense to me, since E is proportional to the inverse radius squared.

I have no way to see if this problem is correct, as it comes from a textbook that only shows answers to odd problems and this is an even problem. Thank for any help in advance!

2. Jun 24, 2017

### Staff: Mentor

That would be correct.

Your Eq (2) is only valid within the volume that contains the charge. (For r <= R, the radius of the charged sphere.)

3. Jun 24, 2017

### Chemmjr18

Thanks! I just to make sure I'm understanding this topic correctly. Anytime we have a uniform charge distribution, we can imaginarily encompass some or all of that charge distribution in a Gaussian shape and subsequently use Gauss's law to find the electric field at some point relative to the charge distribution? Is it really that simple, or am I just over-simplifying it? Also, if the charge distribution is not uniform (I think this would mean the E-field varies) we have to integrate φ=∫E(x,y)⋅cosθ⋅dA?

4. Jun 24, 2017

### Staff: Mentor

Yes, it really is that simple. Assuming you have the needed symmetry. If you do not have the symmetry, applying Gauss' law might involve integrations that you won't be able to simply calculate.

5. Jun 24, 2017

### Chemmjr18

Got it. Thanks again!