Two parallel plates, each of area 5.00 cm 2 , are separatedby 5.00 mm with purif
ID: 1739697 • Letter: T
Question
Two parallel plates, each of area 5.00 cm2, are separatedby 5.00 mm with purifiednonconducting water between them. A voltage of 9.00 V is applied between the plates.Calculate the following. (a) themagnitude of the electric field between the platesN/C
(b) the charge stored on each plate
nC
(c) the charge stored on each plate if the water is removed andreplaced with air
nC (a) themagnitude of the electric field between the plates
N/C
(b) the charge stored on each plate
nC
(c) the charge stored on each plate if the water is removed andreplaced with air
nC
Explanation / Answer
Given that The area of each plate is (A) = 5.00cm2 = 5.00*10-4m2 = 0.0005m2 = 0.0005m2 Separation between the plates (d) =5.0mm =0.005m The voltage between the plates (V) =9.00V a) The magnitude of the electric field between the platesis E =V/d =9.00V/0.005m =1800V/m b) The charge stored on each plate is Q= CV The capacitance of the capacitor is C =0A/d = (8.85*10-12)( 0.0005m2)/0.005m = 0.885*10-12F =0.885pF Then the charge is Q =CV =(0.885*10-12F)(9.00V) =7.965*10-12C = 0.007965nC c) The capacitance of the capacitor when stored on theplate by water then C =k0A/d =(80.4)(8.85*10-12)( 0.0005m2)/0.005m = (80.4)0.885*10-12F =71.154pF Now when the capaccitor is stored with air then C = k0A/d = (1)(8.85*10-12)( 0.0005m2)/0.005m =0.885*10-12F = 0.885pF Then the charge is Q =CV =(0.885*10-12F)(9.00V) =7.965*10-12C =0.007965nC = 0.007965nC c) The capacitance of the capacitor when stored on theplate by water then C =k0A/d =(80.4)(8.85*10-12)( 0.0005m2)/0.005m = (80.4)0.885*10-12F =71.154pF Now when the capaccitor is stored with air then C = k0A/d = (1)(8.85*10-12)( 0.0005m2)/0.005m =0.885*10-12F = 0.885pF Then the charge is Q =CV =(0.885*10-12F)(9.00V) =7.965*10-12C =0.007965nC =(80.4)(8.85*10-12)( 0.0005m2)/0.005m = (80.4)0.885*10-12F =71.154pF Now when the capaccitor is stored with air then C = k0A/d = (1)(8.85*10-12)( 0.0005m2)/0.005m =0.885*10-12F = 0.885pF Then the charge is Q =CV =(0.885*10-12F)(9.00V) =7.965*10-12C =0.007965nC = (1)(8.85*10-12)( 0.0005m2)/0.005m =0.885*10-12F = 0.885pF Then the charge is Q =CV =(0.885*10-12F)(9.00V) =7.965*10-12C =0.007965nC Then the charge is Q =CV =(0.885*10-12F)(9.00V) =7.965*10-12C =0.007965nC =0.007965nCRelated Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.