29: Problem 29.29

INTRO: A 1.50-cm-diameter parallel-plate capacitor with a spacing of 0.700mm is charged to 500V .

PART A: 
What is the total energy stored in the electric field?
Express your answer with the appropriate units.

SOLUTION: 
First off I'm going to convert all of the units to SI units
diameter = 1.50 cm = 0.015 m
spacing = 0.700 mm = 7×10^-4 m
The energy stored in a parallel plate capacitor is determined using the following formulas
U = Q²/(2C) = (CV²)/2 = (QV)/2
We can get Q by using the formula V=(Qd)/(ε₀A) ⇒ Q = (Vε₀A)/d
A = πr² = π(diameter/2)² = 1.77×10^-4 m²
plugging all of our values into the formula for Q we get
Q = 500V ⋅ 8.85×10^-12 C/V ⋅ 1.77×10^-4 m² ⋅1/7×10^-4 m
which gives us Q = 1.119×10^-9 C
Plugging this value of Q into the formula for energy we get
U = 1.119×10^-9 C ⋅ 500 V ⋅ 1/2 = 2.798×10^-7 J 

PART B: 
What is the energy density?
Express your answer with the appropriate units.

SOLUTION: 
The formula for energy density is u= U/volume
the volume of the capacitor is A⋅spacing = 1.77×10^-4 m²⋅7×10^-4 m
giving us Vol=1.239×10^-7 m³
plugging that back into the formula for energy density,
u=2.798×10^-7 J / 1.239×10^-7 m³
= 2.258 J/m³