問題詳情

17. Capacitor and dielectric slab. Consider (see Figure 3.) a capacitor with a fixed charge Q but without adielectric slab (vacuum permittivity is ε0). Its energy is denoted as U0 . A dielectric slab of dielectricconstant K(i.e, permittivity becomesKε0) is now inserted into the plates. Find the work W requiredto insert the dielectric between the plates.


(A)


(B) W=U0(K+1)
(C)


(D)

參考答案

答案：[無官方正解]
難度：計算中-1
書單：沒有書單，新增