Energy Density in Electric Field
Energy density is described as the amount of energy stored in a region or system per unit volume. Electric and magnetic fields can store energy. Energy density is denoted by using the letter u.
Consider a parallel plate capacitor of plate area A and plate separation d. Let the charge on the capacitor be Q, then
The electric field in the region between the plates is,
\(E=\frac{\sigma }{{{\varepsilon }_{0}}}=\frac{Q}{A{{\varepsilon }_{0}}}\).
Volume of capacitor (V) = Ad.
Energy stored, \(\left( U \right)=\frac{{{Q}^{2}}}{2C}=\frac{{{Q}^{2}}}{2{{\varepsilon }_{0}}A}={{\left( \frac{Q}{A{{\varepsilon }_{0}}} \right)}^{2}}\frac{1}{2}{{\varepsilon }_{0}}Ad\).
Energy per unit volume, \(\left( u \right)=\frac{U}{V}={{\left( \frac{Q}{A{{\varepsilon }_{0}}} \right)}^{2}}\frac{1}{2}\frac{{{\varepsilon }_{0}}Ad}{Ad}=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}\).
Energy per unit volume (u) = ½ ε₀E².
Although we have proved the aboveresult for a parallel plate capacitor, in general this is true for any kind ofcapacitor or any other kind of electric field. Now, the potential energy of a sphericalcapacitor made of a single sphere using concept of energy density:
Energy density, (u) = ½ ε₀E².
\(dU=udV=\frac{1}{2}{{\varepsilon }_{0}}{{\left( \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{r}^{2}}} \right)}^{2}}4\pi {{r}^{2}}dr\).
\(U=\frac{1}{8\pi {{\varepsilon }_{0}}}{{Q}^{2}}\int\limits_{R}^{\infty }{\frac{dr}{r}=\frac{{{Q}^{2}}}{8\pi {{\varepsilon }_{0}}R}}\).
How to find the Energy density?
Problem: Calculate the energy density of a capacitor if the electric field is 5V/m?
Solution: Given,
Electric field (E) = 5 V/m
ε₀ = 8. 85 x 10¹² F/ m
We know that:
The energy density formula for capacitor:
Energy Density (u) = ½ ε₀E² = ½ x 8.85 x 10⁻¹² x 5² = 1.10 x 10⁻¹⁰
∴ Energy Density (u) = 1.10 x 10⁻¹⁰ FV²/ m³.