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Problem 2: Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q = 25 nC. The area of each plate is A = 0.024 m2. The distance between them is d = 9.8 cm. The plates are in the air. d A +Q -Q Part (a) With the information given, write the equation of the capacitance in terms of 89, A, d. C = ( 20 A )/d v Correct! Part (b) Solve for the numerical value of C in pF. C = 2.17 / Correct! Part (c) With the information given, express the potential difference across the capacitor in terms of Q and C. AV = QC / Correct! Part (d) Solve for the numerical value of AV in V. AV = 11530 / Correct! Part (e) If the charges on the plates are increased to 2Q, what is the value of the capacitance C in pF? C = Part (f) If A is then increased to 2.4, d is decreased to d/4, what is the value of the capacitance C in pF? C = Part (g) What is the new value of potential difference AV in V with the original charge Q, given the values for A and d from part (f)? AV =

Problem 2: Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q = 25 nC. The area of each plate is A = 0.024 m2. The distance between them is d = 9.8 cm. The plates are in the air. d A +Q -Q Part (a) With the information given, write the equation of the capacitance in terms of 89, A, d. C = ( 20 A )/d v Correct! Part (b) Solve for the numerical value of C in pF. C = 2.17 / Correct! Part (c) With the information given, express the potential difference across the capacitor in terms of Q and C. AV = QC / Correct! Part (d) Solve for the numerical value of AV in V. AV = 11530 / Correct! Part (e) If the charges on the plates are increased to 2Q, what is the value of the capacitance C in pF? C = Part (f) If A is then increased to 2.4, d is decreased to d/4, what is the value of the capacitance C in pF? C = Part (g) What is the new value of potential difference AV in V with the original charge Q, given the values for A and d from part (f)? AV =

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Problem 2: Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q = 25 nC. The area of each plate is A = 0.024 m2. The distance between them is d = 9.8 cm. The plates are in the air. d. A +Q -Q Part (a) With the information given, write the equation of the capacitance in terms of ɛ9, A, d. C = (20 A)/d v Correct! Part (b) Solve for the numerical value of C in pF. C= 2.17 V Correct! Part (c) With the information given, express the potential difference across the capacitor in terms of Q and C. AV = QC / Correct! Part (d) Solve for the numerical value of AV in V. AV = 11530 / Correct! Part (e) If the charges on the plates are increased to 2Q, what is the value of the capacitance C in pF? C = Part (f) If A is then increased to 24, d is decreased to d4, what is the value of the capacitance C in pF? C = Part (g) What is the new value of potential difference AV in V with the original charge Q. given the values for A and d from part (f)? AV =
Transcribed Image Text:Problem 2: Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q = 25 nC. The area of each plate is A = 0.024 m2. The distance between them is d = 9.8 cm. The plates are in the air. d. A +Q -Q Part (a) With the information given, write the equation of the capacitance in terms of ɛ9, A, d. C = (20 A)/d v Correct! Part (b) Solve for the numerical value of C in pF. C= 2.17 V Correct! Part (c) With the information given, express the potential difference across the capacitor in terms of Q and C. AV = QC / Correct! Part (d) Solve for the numerical value of AV in V. AV = 11530 / Correct! Part (e) If the charges on the plates are increased to 2Q, what is the value of the capacitance C in pF? C = Part (f) If A is then increased to 24, d is decreased to d4, what is the value of the capacitance C in pF? C = Part (g) What is the new value of potential difference AV in V with the original charge Q. given the values for A and d from part (f)? AV =
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