Question

A parallel-plate capacitor of capacitance Co, plate area A, spacing d is charged to voltage V. and then disconnected from the charging battery. A slab with dielectric constant K and thickness d/2 is thrust into the capacitor, as shown in the figure below; the slab is exactly halfway between the plates. к (a) What is the new capacitance in terms of Co? (b) What is the ratio of the stored energy before to that after the slab is inserted (U/0.)? (c) How much work is done on the slab as it is inserted? (d) Is the slab sucked in or must it be pushed in?

Answer 1

: ca al mew Capacitance & Co= dtdt. 46 2Gok 9 it)" SPAD . b) stored energy = t co2=0 Of 9 remain V = potential = I fame in this ; 'V' = 9 - į process new potention Potentica Po mew stored energy - Vo = { clvia = ICO V2 v2 (1++] = cv2 [[++))

yx (c) work done :-) 3 3 W=AU W = Uo-U = Lcv2 [+(+#)] - Love = Love [ 4(144) --] JL.CUPS ke 11 11. (d) when we insert dielectric in a charged Capacitor then dielectric is attracted by the Ceipacitor which makes dielectric being "sucked up" by the capacitori . & Heat produced when dielectric inserted im a Capacitor.