2 An air-filled parallel-plate capacitor has plates of area
3 $2.30\U{cm$^2$}$ separated by $1.50\U{mm}$. \Part{a} Find the value
4 of its capacitance. The capacitor is connected to a $12.0\U{V}$
5 battery. \Part{b} What is the charge on the capacitor? \Part{c} What
6 is the magnitude of the uniform electric field between the plates?
11 Using Gauss' law, we can find the electric field from an infinite
12 plate with charge density $\sigma$.
14 \Phi_E &= \oint_S \vect{E}\cdot\vect{dA} = 2EA \\
15 &= \frac{q_\text{in}}{\kappa\varepsilon_0}
16 = \frac{\sigma A}{\kappa\varepsilon_0} \\
17 E &= \frac{\sigma}{2\kappa\varepsilon_0} \;.
20 The capacitance of a ciruit element is defined by $C\equiv q/V$, where
21 the potential difference $V$ is measured with $+q$ on one side of the
22 element and $-q$ on the other. The electric field between the plates
25 E = 2\frac{\sigma}{2\kappa\varepsilon_0}
26 = \frac{q}{\kappa\varepsilon_0 A} \;,
28 and the potential difference is
30 V = Ed = \frac{qd}{\kappa\varepsilon_0 A} \;.
33 Now that we have a formula for $V(q)$, we can plug into the capacitor
34 equation to find the capacitance.
36 C \equiv \frac{q}{V} = \frac{\kappa\varepsilon_0 A}{d}
37 = \frac{1.00 \cdot 8.85\E{-12}\U{C$^2$/N$\cdot$m$^2$}
38 \cdot 2.30\E{-4}\U{m$^2$}}
40 = \ans{1.36\E{-12}\U{F}} = \ans{1.36\U{pF}} \;.
42 As you'd expect, $C$ increases with increasing $A$, and decreases with
46 Just plugging into the capacitance equation,
48 q = CV = 1.36\E{-12}\U{F} \cdot 12.0\U{V}
49 = \ans{1.63\E{-11}\U{C}} = \ans{16.3\U{pC}} \;.
53 Plugging into my earlier equation for electric field,
55 E = \frac{V}{d} = \frac{12.0\U{V}}{1.50\E{-3}\U{m}}
56 = \ans{8.00\U{kV/m}} \;.