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Problem 1: Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q= 37 nC. The area of each plate is A = 0.015 m2. The distance between them is d= 6.2 cm. The plates are in the air. d. A +Q Part (a) With the information given, write the equation of the capacitance in terms of ɛ0, 4, d. C= 20 A/d v Correct! Part (b) Solve for the numerical value of C in pF. C= 2.14 v Correct! Part (c) With the information given, express the potential difference across the capacitor in terms of Q and C. AV = QC v Correct! Part (d) Solve for the numerical value of AV in V. AV = 1.727 * 10(- 8) AV = 1.727E-8 X Attempts Remain Part (e) If the charges on the plates are increased to 20, what is the value of the capacitance C in pF? C= 2.14 v Correct! Part (f) If A is then increased to 24, d is decreased to d/4, what is the value of the capacitance C in pF? / Correct! C= 17 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 1: Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q= 37 nC. The area of each plate is A = 0.015 m2. The distance between them is d= 6.2 cm. The plates are in the air. d. A +Q Part (a) With the information given, write the equation of the capacitance in terms of ɛ0, 4, d. C= 20 A/d v Correct! Part (b) Solve for the numerical value of C in pF. C= 2.14 v Correct! Part (c) With the information given, express the potential difference across the capacitor in terms of Q and C. AV = QC v Correct! Part (d) Solve for the numerical value of AV in V. AV = 1.727 * 10(- 8) AV = 1.727E-8 X Attempts Remain Part (e) If the charges on the plates are increased to 20, what is the value of the capacitance C in pF? C= 2.14 v Correct! Part (f) If A is then increased to 24, d is decreased to d/4, what is the value of the capacitance C in pF? / Correct! C= 17 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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Problem 1: Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q = 37 nC. The area of each plate is A = 0.015 m2. The distance between them is d= 6.2 cm. The plates are in the air. d A +Q Part (a) With the information given, write the equation of the capacitance in terms of ɛ9, 4, d. C= 20 A/d vCorrect! Part (b) Solve for the numerical value of C in pF. / Correct! C= 2.14 Part (c) With the information given, express the potential difference across the capacitor in terms of Q and C. V Correct! AV = O/C Part (d) Solve for the numerical value of AV in V. AV = 1.727 * 10(- 8 ) AV = 1.727E-8 X Attempts Remain Part (e) If the charges on the plates are increased to 20, what is the value of the capacitance C in pF? V Correct! C= 2.14 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? / Correct! C = 17 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 = sin() cos() tan() 8 HOME cotan() asin() acos() 6. atan() acotan() sinh() 1 2 3 cosh() tanh() cotanh() - END O Degrees O Radians VOI BACKSPACE DE CLEAR
Transcribed Image Text:Problem 1: Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q = 37 nC. The area of each plate is A = 0.015 m2. The distance between them is d= 6.2 cm. The plates are in the air. d A +Q Part (a) With the information given, write the equation of the capacitance in terms of ɛ9, 4, d. C= 20 A/d vCorrect! Part (b) Solve for the numerical value of C in pF. / Correct! C= 2.14 Part (c) With the information given, express the potential difference across the capacitor in terms of Q and C. V Correct! AV = O/C Part (d) Solve for the numerical value of AV in V. AV = 1.727 * 10(- 8 ) AV = 1.727E-8 X Attempts Remain Part (e) If the charges on the plates are increased to 20, what is the value of the capacitance C in pF? V Correct! C= 2.14 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? / Correct! C = 17 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 = sin() cos() tan() 8 HOME cotan() asin() acos() 6. atan() acotan() sinh() 1 2 3 cosh() tanh() cotanh() - END O Degrees O Radians VOI BACKSPACE DE CLEAR
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