# Potential across a conducting sphere surrounded by an insulator

• arturo
In summary: I re-ran through the calculations and got .47 (.465).In summary, 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. The thickness of the dielectric must be 1.32*10-2 meters.
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.

arturo said:
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.

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 ...

arturo said:
R = 2.7 ...
OK. What did you get for the thickness of the dielectric?

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:
Yes. I got about .47 m.

arturo said:
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.

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

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

## 1. What is the potential across a conducting sphere surrounded by an insulator?

The potential across a conducting sphere surrounded by an insulator is the difference in electric potential between the inner and outer surfaces of the sphere. It represents the amount of work needed to move a unit positive charge from one surface to the other.

## 2. How is the potential across a conducting sphere affected by the presence of an insulator?

The presence of an insulator surrounding a conducting sphere creates an electric field within the insulator that induces a charge on the surface of the sphere. This induced charge causes a change in the potential across the sphere, which can be calculated using the capacitance of the system.

## 3. What factors influence the potential across a conducting sphere surrounded by an insulator?

The potential across a conducting sphere surrounded by an insulator is influenced by the charge on the sphere, the distance between the inner and outer surfaces of the sphere, the permittivity of the insulator, and the presence of any external electric fields.

## 4. How does the potential across a conducting sphere change with the charge on the sphere?

The potential across a conducting sphere is directly proportional to the charge on the sphere. As the charge increases, the potential also increases. This relationship is described by the equation V = Q/C, where V is the potential, Q is the charge, and C is the capacitance of the system.

## 5. What is the significance of the potential across a conducting sphere surrounded by an insulator?

The potential across a conducting sphere surrounded by an insulator is an important concept in understanding the behavior of electric fields and charges in this type of system. It can be used to calculate the energy stored in the system and the force on the charges, and it also plays a role in determining the stability and behavior of the system.

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