A parallel-plate capacitor has a plate separation of 2.10 mm. If the material between the plates is air, what plate area A is required to provide a capacitance of 2.90 pF? A = m²

A parallel-plate capacitor initially has Quartz between its plates and carries charge ±Q. A different dielectric is then inserted between the plates without changing the charge. If the energy stored in the capacitor decreases to 72% of its initial value, what is the value of x for the new dielectric? K =

A capacitor has a charge of 2.0 µC when connected to a 5.5 V battery. How much energy is stored in this capacitor (in µJ)? +

A parallel-plate capacitor with vacuum between the two plates has the plate area (of one plate) being 0.753 m2 and the plate separation being 7.61E-4 m. What is the capacitance of the capacitor (in F)?   A parallel-plate capacitor with vacuum between the two plates has the plate area (of one plate) being 0.753 m2 and the plate separation being 1.33E-4 m. If we use a battery to maintain a constant 9.51 V potential difference between the plates, what is the electric field strength between the two plates (in V/m)?   A parallel-plate capacitor with vacuum between the two plates has the plate area (of one plate) being 0.753 m2 and the plate separation being 1.33E-4 m. If we use a battery to maintain a constant  2.05 V potential difference between the plates, what is the magnitude of charge stored on one of the plates (in C)?   All three of these are part of the same question please answer all 3

A parallel plate capacitor has an area of A = 94 cm2, a capacitance C = 727 pF, and a separation D = 0.68 mm between the plates. What must the dielectric constant κ of the material be between the plates?

A capacitor stores potential energy PE when charged with a voltage of ΔV. If you double thevoltage, how much energy does the capacitor store?

The plates of an empty parallel-plate capacitor of capacitance 5.2 pF are 2 mm apart. What is the area of each plate? A = m²

Two plates of area 865 x 104 m2 each are separated by a distance of 21.7 x 10-2 m. Calculate the capacitance C of the capacitor. C = F

A parallel-plate capacitor has capacitance C0 = 8.00 pF when there is air between the plates. The separation between the plates is 1.20 mm. Express your answer with the appropriate units. Part A: What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00×104 V/m? Part B: A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00×104 V/m?

Two parallel plates having equal and opposite charges with area L x L separation d is filled with a dielectric material as shown in the Figure. The related dielectric constant varies with zask=a+ Here, a 46 cm, b -92 cm, d 0.31 cm and L=1m are given. Find the capacitance of this capacitor in 10-11 Farad. Take o=10-¹¹ C²/Nm². Answer: