# Potential across a conducting sphere surrounded by an insulator

arturo

## Homework Statement

A conducting sphere has a radius of 2.25 m and carries a positive surplus charge of 35.0 mC. A protective layer of barium titanate is applied to the surface of the sphere to make it safe for laboratory workers nearby. Safety considerations dictate that the potential difference between the surface of the conductor and the outside of the nonconductive layer must be 20,000 V.
How thick must the protective layer be?

1200 = dielectric constant of barium titanate.

## Homework Equations

• Gauss' law for dielectrics:
• ∫E⋅da = qfree, enc/(εκ)
• v = -∫E⋅dl

## The Attempt at a Solution

I began by creating a spherical gaussian surface around the object:
∫E⋅da = qfree, enc/(εκ)
EA = qfree, enc/(εκ)
E4πr2 = qfree, enc/(εκ)
E = qfree, enc/(εκ4πr2)

Then I set up the integral for V:
V = -∫E⋅dl
V = -∫qfree, enc/(εκ4πr2) dr

We want the voltage drop from the surface (r = 2.25m) to some R where ΔV = 20,000.
ΔV = -∫2.25R qfree, enc/(εκ4πr2) dr
ΔV =qfree, enc/(εκ4π*2.25)- qfree, enc/(εκ4πR])
20,000 = qfree, enc/(εκ4π*2.25)- qfree, enc/(εκ4πR])
R = 1.32*10-2

Which is incorrect. Any pointers in the right direction would be appreciated.

Homework Helper
Gold Member
ΔV =qfree, enc/(εκ4π*2.25)- qfree, enc/(εκ4πR])
20,000 = qfree, enc/(εκ4π*2.25)- qfree, enc/(εκ4πR])
I think these equations are OK. But I don't get your answer for R. Try solving the first equation listed here for 1/R before substituting any numbers.

arturo
Okay, one thing I noticed was I accidentally use the wrong multiplier to convert the charge to coulombs, but I'm still incorrect.
20,000 = q/(4πε0κ) [1/2.25 - 1/R]
.076 = [1/2.25 - 1/R]
.368 = 1/R
R = 2.7 ...

Homework Helper
Gold Member
R = 2.7 ...
OK. What did you get for the thickness of the dielectric?

arturo
Sorry I got caught up in noticing my mistake with the power of ten that I didn't do it in variables.
ΔV⋅κ⋅4πε/q = [ 1/Rinital - 1/RFinal
ΔV⋅κ⋅4πε/q + 1/Rinital = 1/RFinal

Your comment + typing it out like this I realize that the R I'm solving for is the final radius not the thickness.
so Thickness = 2.7-2.25 = .45 meters
Does this sound correct?

Edit:
Fixed a typo

Last edited:
Homework Helper
Gold Member
Yes. I got about .47 m.

Homework Helper
Gold Member
ΔV⋅κ⋅4πε/q = [ 1/Rinital - 1/RFinal
ΔV⋅κ⋅4πε/q - 1/Rinital = 1/RFinal
I believe you have a sign error in the second equation. Maybe just a typo.

arturo
Re- ran through the calculations got .47 (.465).
Thank you so much for your help. You guys are always great.

Edit:
I believe you have a sign error in the second equation. Maybe just a typo.
yeah, just a typo