Item 18 Two 12.0 cm-diameter electrodes 0.50 cm apart form a parallel-plate capa

Question

Item 18 Two 12.0 cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 V battery Part A What is the charge on each electrode while the capacitor is attached to the battery? Enter your answers numerically separated by a comma. Submit My Answers Give Up Part B What is the electric field strength inside the capacitor while the capacitor is attached to the battery? Express your answer with the appropriate units. 4 E. Value Units Submit My Answers Give Up Part C What is the potential difference between the electrodes while the capacitor is attached to the battery? Express your answer with the appropriate units.

Explanation / Answer

D =12 cm , r = 6 cm , d = 0.5 cm

V = 14 V

capacitance of parallel plate capacitor

C = Aeo/d

C = 3.14*0.06^2*8.85*10^-12/0.005

C = 2*10^-11 F

Part A:

Q = VC

Q = 14*2*10^-11 = 2.8*10^-10 C

q1 = 2.8*10^-10 C, q2 = -2.8*10^-10 C

Part B:

E = Q/Aeo

E = 2.8*10^-10/(3.14*0.06^2*8.85*10^-12)

E = 2799 V/m

PartC:

V = 14 V

Part D: d = 1.7 cm

C = 3.14*0.06^2*8.85*10^-12/0.017

C = 5.885*10^-12 F

Q = 5.88*10^-12*14 = 8.232*10^-11 C

q1 = 8.232*10^-11 C, q2 = - 8.232*10^-11 C

Part E:

E = 2799 V/m

(f) V = E*d = 2799*0.017

V = 47.583 V

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