Capacitance and charge of a capacitor

A parallel-plate capacitor has plates of area $0{,}010$ m$^2$ separated by $1{,}0$ mm of vacuum. Calculate the capacitance. If it is charged to $100$ V, what charge is stored?

Solution

Setup. The capacitance of a parallel-plate capacitor with vacuum between the plates is $C = \varepsilon_0 \frac{A}{d}$ while the charge stored at a given potential difference is $Q = C\,V$. Data. $A = 0{,}010 \ \text{m}^2$, $d = 1{,}0 \text{ mm} = 1{,}0\cdot 10^{-3}$ m, $\varepsilon_0 = 8{,}85\cdot 10^{-12} \ \text{F/m}$, $V = 100$ V. Capacitance. $C = (8{,}85\cdot 10^{-12})\,\frac{0{,}010}{1{,}0\cdot 10^{-3}} = (8{,}85\cdot 10^{-12})\cdot 10$ $\ev{C \approx 88{,}5 \ \text{pF}}$ Charge. $Q = C\,V = (8{,}85\cdot 10^{-11})(100)$ $\ev{Q \approx 8{,}85\cdot 10^{-9} \ \text{C} = 8{,}85 \ \text{nC}}$

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Topics: Electric field and potential Concepts: Capacitance and capacitor · Electric charge Skills: Superposition principle Methods: Gradient method Objects: Parallel-plate capacitor