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.
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 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.
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 is 13 pF. Determine the seperation between the plates of the capacitor. (10 points)
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
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 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...
A parallel-plate capacitor has plates of area 3.55×10−4 m2 A) What plate separation is required if the capacitance is to be 1490 pF ? Assume that the space between the plates is filled with air? (Dielectric constant for air is 1.00059) B) What plate separation is required if the capacitance is to be 1490 pF ? Assume that the space between the plates is filled with paper. (Dielectric constant for paper is 3.7)
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...
Question 4 1 pts A parallel-plate capacitor of capacitance C has the area of plates A and distance between plates d. Capacitor is connected to the battery of voltage V, fully charged and then disconnected. A slab of dielectric material with dielectric constant 4 is then inserted into capacitor, completely filling region between plates. Electric field between plates of capacitor: decreases twice stays the same increases twice decreases 4 times increases 4 times
A parallel-plate air capacitor with a capacitance of 244 pF has a charge of magnitude 0.137 mu C on each plate. The plates have a separation of 0.265 mm. What is the potential difference between the plates? What is the area of each plate? Use 8.85 times 10^-12 F/m for the permittivity of free space. What is the electric field magnitude between the plates? What is the surface-charge density on each plate?