Part A)
First of all, let's convert the radii of the inner and the outer sphere:
![r_A = 11.0 cm = 0.110 m](https://tex.z-dn.net/?f=r_A%20%3D%2011.0%20cm%20%3D%200.110%20m)
![r_B = 16.5 cm=0.165 m](https://tex.z-dn.net/?f=r_B%20%3D%2016.5%20cm%3D0.165%20m)
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
![C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=](https://tex.z-dn.net/?f=C%3D4%20%5Cpi%20%5Cepsilon%20_0%20%20%5Cfrac%7Br_A%20r_B%7D%7Br_B-%20r_A%7D%3D4%5Cpi%288.85%20%5Ccdot%2010%5E%7B-12%7DC%5E2m%5E%7B-2%7DN%5E%7B-1%7D%29%20%5Cfrac%7B%280.110m%29%280.165m%29%7D%7B0.165m-0.110m%7D%3D)
![=3.67\cdot 10^{-11}F](https://tex.z-dn.net/?f=%3D3.67%5Ccdot%2010%5E%7B-11%7DF)
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
![Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C](https://tex.z-dn.net/?f=Q%3DCV%3D%283.67%5Ccdot%2010%5E%7B-11%7DF%29%28100%20V%29%3D3.67%5Ccdot%2010%5E%7B-9%7DC)
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
![E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}](https://tex.z-dn.net/?f=E%5Ccdot%20%284%20%5Cpi%20r%5E2%29%20%3D%20%20%5Cfrac%7BQ%7D%7B%5Cepsilon%20_0%7D%20)
from which
![E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}](https://tex.z-dn.net/?f=E%28r%29%20%3D%20%20%5Cfrac%7BQ%7D%7B4%20%5Cpi%20%5Cepsilon_0%20r%5E2%7D)
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
![E(0.111m)=2680 N/C](https://tex.z-dn.net/?f=E%280.111m%29%3D2680%20N%2FC)
And so, the energy density at r=0.111 m is
![U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3](https://tex.z-dn.net/?f=U%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cepsilon%20_0%20E%5E2%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%288.85%5Ccdot%2010%5E%7B-12%7DC%5E2m%5E%7B-2%7DN%5E%7B-1%7D%29%282680%20N%2FC%29%5E2%3D3.17%20%5Ccdot%2010%5E%7B-5%7DJ%2Fm%5E3)
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
![Q=3.67 \cdot 10^{-9}C](https://tex.z-dn.net/?f=Q%3D3.67%20%5Ccdot%2010%5E%7B-9%7DC)
. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
![E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C](https://tex.z-dn.net/?f=E%280.164%20m%29%3D%20%5Cfrac%7BQ%7D%7B4%20%5Cpi%20%5Cepsilon_0%20r%5E2%7D%3D1228%20N%2FC%20)
And therefore, the energy density at this distance from the center is