28: Problem 28.19

INTRO:
A 3.4-cm-diameter parallel-plate capacitor has a 1.6 mm spacing. The electric field strength inside the capacitor is 8.0×10⁴ V/m .
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PART A:
What is the potential difference across the capacitor?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

My favorite first step,​

Givens/ conversions:
diameter ≡ d = 3.4 cm = 0.034 m
spacing ≡ s = 1.6 mm = 0.0016 m
Electric field inside ≡ E = 8.0×10⁴ V/m

using equation 28.5, for the electric potential inside a parallel-plate capacitor: V = Es

so, V = (8.0×10⁴ V/m)(0.0016 m) = 128 V

the frustrating kicker here, it got me anyway, is that the instructions say to express your solution to two significant figures. 
128 has 3 sigfigs. annoying. You have to round up to 130...

⇒ ΔV_C = 130 V

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PART B:
How much charge is on each plate?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
An electric field is considered to be a surface charge density (with units of C/m²) divided by the vacuum permittivity constant (with units C/Vm). This means that the surface charge density, η, can be calculated by multiplying the electric field by the vacuum permittivity constant. 

η=E⋅ε₀​

the surface charge density is basically the charge distributed over an area, or η = Q/A
therefore, to solve for the charge, 

Q = ηA = E⋅ε₀⋅A
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The area is just the area of the circular plate, or A = πr² = π/4 ⋅ d²

∴ Q = E⋅ε₀⋅π/4⋅d²

= (8.0×10⁴ V/m)(π/4)(8.854×10^-12 C/Vm)(0.034 m)²
⇒ Q = 6.431×10^-10 C​

but don't forget sigfigs... it only wants two...

⇒ Q = 6.4×10^-10 C​