What is the capacitance of a Ruby mica capacitor with an area of 2 by 1...
What is the capacitance of a Ruby mica capacitor with an area of 2 by 1 cm with a distance between the plates of 4 mm? Write your answer in picofarads. For example, if you got 3.18 x 10-10 F, that would be 318 pF.
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What is the capacitance of a Ruby mica capacitor with an area of
2 by 1...
What is the capacitance of a Ruby mica capacitor with an area of 3 by 4...
What is the capacitance of a Ruby mica capacitor with an area of 3 by 4 cm with a distance between the plates of 2 mm? Write your answer in pico farads. For example, if you got 3.18 x 10-10 F, that would be 318 pF.
Compute the capacitance of a parallel-plate capacitor with plates 5 cm by 4 cm, separation of 2 mm, and a mica dielectric
Compute the capacitance of a parallel-plate capacitor with plates 5 cm by 4 cm, separation of 2 mm, and a mica dielectric 48.7 PF 68.2 pF 119.8 pF 169.5 pF
2. A parallel plate capacitor has an area of 80cm and the capacitance of the capacitor...
2. A parallel plate capacitor has an area of 80cm and the capacitance of the capacitor is 13 pF. Determine the seperation between the plates of the capacitor. (10 points)
An air-filled parallel-plate capacitor has plates of area 2.80 cm^2 separated by 2.50 mm. The capacitor...
An air-filled parallel-plate capacitor has plates of area 2.80 cm^2 separated by 2.50 mm. The capacitor is connected to a(n) 17.0 V battery. Find the value of its capacitance. _____ pF What is the charge on the capacitor? ______ pC What is the magnitude of the uniform electric field between the plates ___________ N/C
1. A parallel plate capacitor has an area of 1,000 cm ^ 2 and a distance...
1. A parallel plate capacitor has an area of 1,000 cm ^ 2 and a distance between the plates of 1 cm. A dielectric material is inserted and its new capacitance is calculated to be 1.77 x 10 ^ -10 F. Determine the value of the dielectric constant. a. 2 b. 2.25 c. 1.5 d. 0.75
A capacitor has plates with area A = 0.040 m^2, separation d = 1.6 mm, with...
A capacitor has plates with area A = 0.040 m^2, separation d =
1.6 mm, with a piece of material in between. Its capacitance is 5.0
x 10^-10 F. What is the dielectric constant of the material K =
?
A capacitor has plates with area A = 0.040 m^2, separation d = 1.6 mm, with a piece of material in between. Its capacitance is 5.0 x 10^-10 F. What is the dielectric constant of the material K = ?...
Question An air-filled parallel-plate capacitor is to have a capacitance of 0.6 F If the distance...
Question An air-filled parallel-plate capacitor is to have a capacitance of 0.6 F If the distance between the plates is 0.8 mm, calculate the required surface area of each plate. The permittivity of free space is 8.8542 x 10-12 C2/N m2. Answer in units of km2
A parallel-plate capacitor has plates of area 2.88×10−4 m2 . Part A. What plate separation is...
A parallel-plate capacitor has plates of area 2.88×10−4 m2 . Part A. What plate separation is required if the capacitance is to be 1660 pF ? Assume that the space between the plates is filled with air? (Dielectric constant for air is 1.00059) Express your answer using three significant figures.d= ? μm Part B What plate separation is required if the capacitance is to be 1660 pF ? Assume that the space between the plates is filled with paper. (Dielectric...
Capacitor calculation a) Calculate the capacitance of a parallel-plate capacitor whose plates are made of two...
Capacitor calculation a) Calculate the capacitance of a parallel-plate capacitor whose plates are made of two different sizes. One plate has a radius of 10 cm and the other plate has 12 cm and is separated by 0.75 mm air gap. b) What is the charge on each plate if a 12-V battery is connected across the two plates? c) What is the electric field between the plates? d) Estimate the area of the plates needed to achieve a capacitance...
Keep getting it wrong because I need the answer in picofarads (pF) 1. 0/2 points Previous...
Keep getting it wrong because I need the answer in picofarads
(pF)
1. 0/2 points Previous Answers PSE6 26.AE.01. Example 26.1 Parallel-Plate Capacitor Problem A parallel-plate capacitor has an area A = 1.90 X 10^4 m- and a plate separation d = 1.26 mm. Find its capacitance. Strategy Use Equation 26.3. €0. Solution From Equation 26.3, we find the following. (1.90 x 10-4 m² = (8.85 x 10-12 C2/N. 1.26 x 10-3 m X PF Your response differs significantly from...
2. A parallel plate capacitor has an area of 80cm and the capacitance of the capacitor is 13 pF. Determine the seperation between the plates of the capacitor. (10 points)
An air-filled parallel-plate capacitor has plates of area 2.80 cm^2 separated by 2.50 mm. The capacitor is connected to a(n) 17.0 V battery. Find the value of its capacitance. _____ pF What is the charge on the capacitor? ______ pC What is the magnitude of the uniform electric field between the plates ___________ N/C
A capacitor has plates with area A = 0.040 m^2, separation d =
1.6 mm, with a piece of material in between. Its capacitance is 5.0
x 10^-10 F. What is the dielectric constant of the material K =
?
A capacitor has plates with area A = 0.040 m^2, separation d = 1.6 mm, with a piece of material in between. Its capacitance is 5.0 x 10^-10 F. What is the dielectric constant of the material K = ?...
Question An air-filled parallel-plate capacitor is to have a capacitance of 0.6 F If the distance between the plates is 0.8 mm, calculate the required surface area of each plate. The permittivity of free space is 8.8542 x 10-12 C2/N m2. Answer in units of km2
Capacitor calculation a) Calculate the capacitance of a parallel-plate capacitor whose plates are made of two different sizes. One plate has a radius of 10 cm and the other plate has 12 cm and is separated by 0.75 mm air gap. b) What is the charge on each plate if a 12-V battery is connected across the two plates? c) What is the electric field between the plates? d) Estimate the area of the plates needed to achieve a capacitance...
Keep getting it wrong because I need the answer in picofarads
(pF)
1. 0/2 points Previous Answers PSE6 26.AE.01. Example 26.1 Parallel-Plate Capacitor Problem A parallel-plate capacitor has an area A = 1.90 X 10^4 m- and a plate separation d = 1.26 mm. Find its capacitance. Strategy Use Equation 26.3. €0. Solution From Equation 26.3, we find the following. (1.90 x 10-4 m² = (8.85 x 10-12 C2/N. 1.26 x 10-3 m X PF Your response differs significantly from...
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