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?
Express your answer using two significant digits.
1200 = dielectric constant of barium titanate.
- 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.