Verizon LTE 5:33 PM a spock.physast.uga.edu Anna K. Sims ? Student PHYS 1212: Pr
ID: 2303687 • Letter: V
Question
Verizon LTE 5:33 PM a spock.physast.uga.edu Anna K. Sims ? Student PHYS 1212: Principles of Summer 2018 Main Menu Contents Grades Syllabus , Pond Course Contents » » Problem Set #5 » (2) Timer Notes Evaluate Feedback ? Print Info A parallel-plate capacitor has an area of 4.50 cm, and the plates are separated by 1.19 mm with air between them. It stores a charge of 355 pC. What is the potential difference across the plates of the capacitor? Submit Answer Tries 0/8 What is the magnitude of the uniform electric field in the region between the plates? Submit AnswerTries 0/8 Post Discussion Send FeedbackExplanation / Answer
Given,
Area of capacitor, A = 4.5 cm^2 = 4.5*10^-4 m^2
Seperation, d = 1.19 mm = 1.19*10^-3 m
Permitivity, ?0 = 8.85*10^-12 C^2/N.m^2
Charge, Q = 355*10^-12 C
Capacitance, C = ?0 A / d
= (8.85*10^-12 * 4.5*10^-4) / (1.19*10^-3)
= 3.348*10^-12 F
potential difference,V = Q /C
= 355 *10^-12 C / 3.348*10^-12 F
= 106.03 V (Ans)
b ) Electric field, E = V / d
= 106.03 / 1.19*10^-3 m
= 89.1 * 10^3 N /C (Ans)
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