Parallel-plate capacitor: capacitance, charge and energy

Problem

A parallel-plate capacitor has plates of area $A = 200\;\text{cm}^2$ separated by an air gap of thickness $d = 1{,}0\;\text{mm}$. It is charged by connecting it to a $V = 12\;\text{V}$ battery, then disconnected. Determine the capacitance, the stored charge, and the total electrostatic energy.

Solution

Data in SI units. $A = 200\;\text{cm}^2 = 2{,}0\cdot 10^{-2}\;\text{m}^2$, $d = 1{,}0\cdot 10^{-3}\;\text{m}$, $V = 12\;\text{V}$.

Capacitance. For an air-filled parallel-plate capacitor: $C = \varepsilon_0\,\frac{A}{d} = (8{,}85\cdot 10^{-12})\,\frac{2{,}0\cdot 10^{-2}}{1{,}0\cdot 10^{-3}}$ $\ev{C \approx 1{,}77\cdot 10^{-10}\;\text{F}}$

Stored charge. From the definition of capacitance $Q = CV$: $Q = (1{,}77\cdot 10^{-10})\,(12) = 2{,}1\cdot 10^{-9}\;\text{C}$ $\ev{Q \approx 2{,}1\cdot 10^{-9}\;\text{C}}$

Electrostatic energy. The energy stored in the capacitor: $W = \tfrac12\,C V^2 = \tfrac12\,(1{,}77\cdot 10^{-10})\,(12)^2 = \tfrac12\,(1{,}77\cdot 10^{-10})\,(144)$ $\ev{W \approx 1{,}3\cdot 10^{-8}\;\text{J}}$

Once the battery is disconnected the charge $Q$ stays constant (the capacitor is isolated); the calculated energy is the one stored in the electric field between the plates.

Links

Topics: Campo elettrico e potenziale Concepts: Capacità e condensatore · Energia potenziale elettrostatica Objects: Condensatore piano