Potential Difference problem with sphere and capacitor

[MAC5494]

Homework Statement

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.

Homework Equations

E_sphere = (1/4ε0)*(q/r²)
E_capacitor = Q/A/ε0
ΔV = - (E*ΔL)

The Attempt at a Solution

a)
E_capacitor = 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.

 

[Adithyan]

Please provide the figure.

 

[MAC5494]

I added it.

 

[BvU]

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

 

[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!

 

[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.

 

[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.