Potential Difference problem with sphere and capacitor

1. Feb 13, 2014

MAC5494

1. The problem statement, all variables and given/known data

A thin spherical shell made of plastic carries a uniformly distributed negative charge -6e-10 coulombs (indicated as -Q1 in the diagram). Two large thin disks made of glass carry uniformly distributed positive and negative charges 1.5e-05 coulombs and -1.5e-05 coulombs (indicated as +Q2 and -Q2 in the figure). The radius R1 of the plastic spherical shell is 6 mm, and the radius R2 of the glass disks is 4 meters. The distance d from the center of the spherical shell to the positive disk is 18 mm.

(a) Find the potential difference V1 - V2. Point 1 is at the center of the plastic sphere, and point 2 is just outside the sphere.

(b) Find the potential difference V2 - V3. Point 2 is just below the sphere, and point 3 is right beside the positive glass disk.

Remember that the sign of the potential difference is important.

2. Relevant equations
Esphere = (1/4ε0)*(q/r2)
Ecapacitor = Q/A/ε0
ΔV = - (E*ΔL)

3. The attempt at a solution

a)
Ecapacitor = Q/A/ε0
= (1.5e-5)/(4^2∏)/(8.85e-12) = 3.4e4 N/C

ΔV = - (E*ΔL)
= - (3.4e4 * .006) = -202.32 V/m

b)

I don't understand how to do part b. I really could use some help. Part A is correct.

Last edited: Feb 13, 2014
2. Feb 13, 2014

3. Feb 13, 2014

MAC5494

4. Feb 13, 2014

BvU

Is it unreasonable to think you can separately calculate the potential differences and add them up ?
(and take good care of the signs!)

5. Feb 13, 2014

MAC5494

I've tried that, and it could be that I'm doing it wrong, maybe mixing up signs, but I'll take another look at it!

6. Feb 14, 2014

BvU

Well, if we find something in the category "The attempt at a solution" in the next post, maybe we can help. That's what PFHW is for.

7. Feb 14, 2014

MAC5494

Sorry, I meant to respond again, I figured out the problem. I was using the wrong equation for finding the potential difference due to the sphere.