Question

8. (-/1.25 Points] DETAILS MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A student working in the physics laboratory connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 270 V. Assume a plate separation of d = 1.54 cm and a plate area of A = 25.0 cm. When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in PC) before and after the capacitor is submerged. (Enter the magnitudes.) before pc after pC (b) Determine the capacitance (in F) and potential difference in V) after immersion. CE - F AVE V (c) Determine the change in energy (in nd) of the capacitor. AU = | n] (d) What If? Repeat parts (a) through (C) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 270 V potential difference. Calculate the charge on the plates (in PC) before and after the capacitor is submerged. (Enter the magnitudes.) before pc after pc Determine the capacitance (in F) and potential difference in V) after immersion. CE = AV= V F Determine the change in energy (in n)) of the capacitor. ΔU - n]

Answer 1

d = 1.54cm AE 25.0cm² (= EA 8.85x10 -12 x 25x10-4 = 1.43x10-12 F d 1.54 x 10-2 = 1.43PF a) 9; = 1.43x270 = 38611 PC Of = 386.1 PC as battery is removed b) Cf=kci = 80 x 1.43x10-12 F F = 114.4 8x10 12 F Avfa Qf 386-1410-12 - 3.375 v Cf 114.4 x 10-12 ΔU - 2 2Cf ( 20i 2 Cf Ci 2 (31061810 114.4x10-12 1.43x10-12 51.472 X10 -9 J= 51.472 nJ (i= 386.1 PC Of = CF V = 80 x 386.1= 30880 PC (f = 114.4x10 10 -12 F AV = 270 v AUR FUL civ> {x79C1V? x792 143_15" = 4117.756 nJ