Question

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm , and the outer sphere has radius 16.5 cm . A potential difference of 100 V is applied to the capacitor.
Part A
What is the energy density at r= 11.1 cm , just outside the inner sphere?
Part B
What is the energy density at r = 16.4 cm , just inside the outer sphere?
How can I solve this if I have no Q!

Answer 1

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
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
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$