Field energy in the capacitor

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Field energy in the capacitor. An air parallel-plate capacitor has square plates of side $\ell=5{,}0\;\text{cm}$ and separation $d=1{,}0\;\text{mm}$. It is charged to $V=100\;\text{V}$.

(a) Calculate the magnitude of the field $E$ between the plates. (b) Calculate the energy density $u=\tfrac12\varepsilon_0 E^2$. (c) Work out the total energy $U=u\cdot\ell^2\cdot d$ and compare it with $\tfrac12 CV^2$: they must match.

Solution

Data in SI units. $\ell = 5{,}0\;\text{cm} = 5{,}0\cdot 10^{-2}\;\text{m}$, $d = 1{,}0\;\text{mm} = 1{,}0\cdot 10^{-3}\;\text{m}$, $V = 100\;\text{V}$.

(a) Field between the plates. The field of a parallel-plate capacitor is uniform and related to the voltage by $E = V/d$: $E = \frac{V}{d} = \frac{100}{1{,}0\cdot 10^{-3}}$ $\ev{E = 1{,}0\cdot 10^{5}\;\text{V/m}}$

(b) Energy density. $u = \tfrac12\,\varepsilon_0\,E^2 = \tfrac12\,(8{,}85\cdot 10^{-12})\,(1{,}0\cdot 10^{5})^2 = \tfrac12\,(8{,}85\cdot 10^{-12})\,(1{,}0\cdot 10^{10})$ $\ev{u \approx 4{,}4\cdot 10^{-2}\;\text{J/m}^3}$

(c) Total energy and check. The volume between the plates is $\ell^2 d = (5{,}0\cdot 10^{-2})^2\,(1{,}0\cdot 10^{-3}) = 2{,}5\cdot 10^{-6}\;\text{m}^3$: $U = u\cdot\ell^2\cdot d = (4{,}4\cdot 10^{-2})\,(2{,}5\cdot 10^{-6})$ $\ev{U \approx 1{,}1\cdot 10^{-7}\;\text{J}}$

Comparison with $\tfrac12 CV^2$. The capacitance is $C = \varepsilon_0\dfrac{\ell^2}{d} = (8{,}85\cdot 10^{-12})\dfrac{2{,}5\cdot 10^{-3}}{1{,}0\cdot 10^{-3}} = 2{,}21\cdot 10^{-11}\;\text{F}$, hence $\tfrac12 C V^2 = \tfrac12\,(2{,}21\cdot 10^{-11})\,(100)^2 = 1{,}1\cdot 10^{-7}\;\text{J}\;\checkmark$ The two methods agree: the energy stored in the capacitor is exactly the energy distributed in the electric field between the plates, with density $u=\tfrac12\varepsilon_0 E^2$.

Links

Topics: Electric field and potential Concepts: Field energy density · Capacitance and capacitor · Electrostatic potential energy Objects: Parallel-plate capacitor