apart.Since the electric field strength in the region between the plates is given by E=o/,the external force can be rewritten as F. (5.3.5) 2 The external force F is independent of d.The total amount of work done externally to separate the plates by a distance d is then 形=∫Es=rd=E4d 2Γ (5.3.6) consistent with Eq.(5.3.3).Because the potential energy of the system is equal to the work done by the external agent,we have that the energy density u=W/Ad=E2/2 In addition,we note that the expression for ug is identical to Eq.(3.4.8)in Chapter 3. Therefore,the electric energy density ug can also be interpreted as electrostatic pressure P Example 5.4:Electric Energy Density of Dry Air The breakdown field strength at which dry air loses its insulating ability and allows a discharge to pass through is E=3x10V/m.At this field strength,the electric energy density is: Nm3x 10*V/m)40 Jm 1 (5.3.7) Example 5.5:Energy Stored in a Spherical Shell Find the energy stored in a metallic spherical shell of radius a and charge O. Solution:The electric field associated of a spherical shell of radius a is(Example 3.3) r,r>a E- (5.3.8) 0, r<a. The corresponding energy density is 1 eE2= 03 4E= 2 (5.3.9) 32π2er4 5-115-11 apart. Since the electric field strength in the region between the plates is given by 0 E = ! / " , the external force can be rewritten as 0 2 ext 2 F E A ! = . (5.3.5) The external force Fext is independent of d . The total amount of work done externally to separate the plates by a distance d is then 2 0 ext ext ext 2 E A W d F d d " ! # = $ = = % & ' ( ) F s ! ! , (5.3.6) consistent with Eq. (5.3.3). Because the potential energy of the system is equal to the work done by the external agent, we have that the energy density 2 ext 0 / / 2 E u =W Ad = ! E . In addition, we note that the expression for E u is identical to Eq. (3.4.8) in Chapter 3. Therefore, the electric energy density E u can also be interpreted as electrostatic pressure P. Example 5.4: Electric Energy Density of Dry Air The breakdown field strength at which dry air loses its insulating ability and allows a discharge to pass through is 6 Eb = 3!10 V/m . At this field strength, the electric energy density is: uE = 1 2 ! 0E2 = 1 2 (8.85 " 10-12 C2 /N #m2 )(3" 106 V/m) 2 = 40 J/m3 . (5.3.7) Example 5.5: Energy Stored in a Spherical Shell Find the energy stored in a metallic spherical shell of radius a and charge Q. Solution: The electric field associated of a spherical shell of radius a is (Example 3.3) E !" = Q 4!" 0 r 2 rˆ, r > a " 0, r < a. # $ % & % (5.3.8) The corresponding energy density is 2 2 0 2 4 0 1 2 32 E Q u E r ! " ! = = , (5.3.9)