# Flux and nonconducting shells

1. Sep 14, 2015

### kilnvzol

1. The problem statement, all variables and given/known data
A charged particle is suspended at the center of two concentric spherical shells that are very thin and made of nonconducting material. Figure (a) shows a cross section. Figure (b) gives the net flux Φ through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere. The scale of the vertical axis is set by Φs = 6.0 × 105 N·m2/C. (a) What is the charge of the central particle? What are the net charges of (b) shell A and(c) shell B? (in µC)
http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c23/pict_23_12.gif

2. Relevant equations
EI = q(enc)

3. The attempt at a solution
For part a. I multiplied (2E5)(8.85E-12) = 1.77µC
For part b. I did -(6E5)(8.85E-12)-1.77E-6 = -7.08µC
For part c. I did (6E5)(8.85E-12)+(7.08E-6)-(1.77E-6) = 10.62µC

I don't know what I did wrong. :(

2. Sep 14, 2015

### BvU

Hello kilnvzol, welcome to PF !

Did you check the scale factor $\Phi_S$ ?

3. Sep 14, 2015

My guess is that you got the first part right but not the other two? The first part seems good, but then you go and start subtracting things. You were correct in counting the number of steps the flux increased immediately outside of the point charge, that was good. But then you notice that the flux drops by 6 steps, and every one of those steps is due to the existence of another shell. So all you need to worry about is the 6 downward steps, you don't have to do any subtraction afterwards. And same goes for the third part.

Does that make any sense?

4. Sep 14, 2015

### kilnvzol

Did I need to multiply (5/6) by (2E5)(8.85E-12)?

I got the first part wrong too.
So its just (6E5)(8.85E-12) without the subtraction?

5. Sep 15, 2015

### BvU

If five divisions represent $6 \times 10^5\$Nm2/C, how much is two divisions ? And six ? And ten ?

6. Sep 15, 2015

Oh, I didn't even catch that. Yeah, kilnvzol, that would be your reason for getting the first part wrong.
Almost, you would still have to account for the value that is actually assumed by one tick mark. But you're right on removing the subtraction. Does it actually make sense why that is?

7. Sep 15, 2015

### kilnvzol

I got it! I forgot to multiply in the scale. XD

Thanks guys!