能量密度公式
可以存储在给定质量的物质或系统中的能量的数量由其能量密度决定。因此,系统或材料的能量密度越高,它可以在其质量中存储的能量就越多。许多不同的物质和技术可用于储存能量。最典型的是,在确定系统中的能量量时,只测量可用或可提取的能量。我们通常在科学方程中计算能量密度。本文将举例讨论能量密度公式。
能量密度
系统中每单位体积的能量总量称为能量密度。例如,每克食物的卡路里数量。低能量密度的食物每克热量更少,可以让你吃更多。它用 U 表示。能量可以储存在磁场和电场中。
能量密度公式
- 电容器或电场的能量密度由下式给出,
UE = (1/2)ε0E2
Where,
- UE = Electrical Energy Density,
- ε0 = Permittivity,
- E = Electric Field.
推导
Energy density = Energy/volume
UE = U / V
Energy = 1/2 [ε0 E2] × Ad
UE = 1/2 [ε0 E2] × Ad / Ad
UE = (1/2)ε0E2
- 磁场或电感器的能量密度由下式给出,
UB = (1/2μ0)B2
Where,
- UB = Magnetic Energy Density,
- μ0 = Permeability
- B = Magnetic Field.
推导
Energy density = Energy/volume
UB = 1/2 [LI2]/Al
Flux = NBA = LI
B = μ0 NI/length
I = B (Length)/ Nμ0
UB = 1/2 {B (Length)/ Nμ0} [NBA]/A (length)
UB = (1/2μ0)B2
磁场和电场都有助于电磁波的能量密度。因此,电场和磁场的总能量密度等于总能量密度。
U = (1/2)ε0E2 + (1/2μ0)B2
示例问题
问题 1:定义能量密度。
回答:
The total quantity of energy in a system per unit volume is known as energy density. Total energy density involves both capacitive energy density and inductive energy density. The formula for the total energy density,
U = (1/2)ε0E2 + (1/2μ0)B2
问题2:磁场或电感的能量密度公式是什么?
答案:
The energy density of a magnetic field or an inductor is given by,
UB = (1/2μ0)B2
Where,
- UB = Magnetic Energy Density,
- μ0 = Permeability
- B = Magnetic Field.
问题 3:计算电场为 E = 12 V/m 的电容器的能量密度。
解决方案:
Given: E = 12 V/m, ε0 = 8.8541 × 10-12 F/m
Since,
UE = (1/2)ε0E2
∴ UE = (1/2) × 8.8541 × 10-12 × 122
∴ UE = (1/2) × 1274.99 × 10-12
∴ UE = 637.495 × 10-12
∴ UE = 6.375 × 10-10 FV2/m3
问题4:在空间的一个区域中,磁场的值为3×10 -2 T。相反,电场的值为3×10 -7 V/m。计算电场和磁场的总能量密度。
解决方案:
Given: B = 3 × 10-2 T, E = 3 × 10-7 V/m, ε0 = 8.8541 × 10-12 F/m, μ0 = 4π × 10-7 NA-2
Since,
U = (1/2)ε0E2 + (1/2μ0)B2
∴ U = ((1/2) × 8.8541 × 10-12 × (3 × 10-7)2) + ((1/(2 × 4π × 10-7)) × (3 × 10-2)2)
∴ U = 3982.5 + 358.1
∴ U = 4340.6 J/m3
问题 5:计算电场为 E = 20 V/m 的电容器的能量密度。
解决方案:
Given : E = 20 V/m, ε0 = 8.8541 × 10-12 F/m
Since,
UE = (1/2)ε0E2
∴ UE = (1/2) × 8.8541 × 10-12 × 202
∴ UE = (1/2) × 3541.64 × 10-12
∴ UE = 1.7 × 10-9 FV2/m3
问题 6:计算磁场为 B = 8 T 的电感器的能量密度。
解决方案:
Given: B = 8 T, μ0 = 4π × 10-7 NA-2
Since,
UB = (1/2μ0)B2
∴ UB = (1/(2 × 4π × 10-7)) × 82
∴ UB = (1/25.12 × 10-7) × 64
∴ UB = 2.5477 × 10-7 J/m3