Use the characteristics of the Coulomb force to explain why capacitance should be proportional to the plate area of a capacitor. Similarly, explain why capacitance should be inversely proportional to the separation between plates.
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Use the characteristics of the Coulomb force to explain why capacitance should be proportional to the plate area of a capacitor. Similarly, explain why capacitance should be inversely proportional to the separation between plates.
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- A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d = 1.47 cm and a plate area of A = 25.0 cm². When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before after AV f (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf= = = (c) Determine the change in energy (in nJ) of the capacitor. AU = nJ AV f (d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 225 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before after…T or F A capacitor is a charge storage device whose capacitance is defined as the ratio of the amount of the potential difference (in volts) across its plates to the charge (in coulombs) on its plates.The volume enclosed by the plates of a parallel plate capacitor is given by the equation Volume = Ad. If one doubles the area, A, of the plates and reduces the plate separation, d, by a factor of two so that the volume remains the same, the capacitance will do which of the following? decrease by a factor of 4 O increase by a factor of 4 increase by a factor of 2 decrease by a factor of 2 O remain constant
- A slab of copper of thickness b = 1.68 mm is thrust into a parallel-plate capacitor of plate area A = 1.96 cm2 and plate separation d = 5.35 mm, as shown in the figure; the slab is exactly halfway between the plates. (a) What is the capacitance after the slab is introduced? (b) If a charge q = 2.68 µC is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? (c) How much work is done on the slab as it is inserted? (d) Is the slab sucked in or must it be pushed in? Copper (a) Number 4.73e-13 Units (b) Number i 0.686 (c) Number i 5.36e-10 Units J (d) sucked inA parallel-plate capacitor with a 0.1-mm air gap has a capacitance of 12 µF. (a) When this capacitor is connected to a battery, the maximum magnitude of charge on either plate is 0.18 mC (millicoulomb). What are the magnitude of the electric field between the plates and the energy density (energy per unit volume) in this region? (b) (c) Now suppose this capacitor is disconnected from the battery, and a slab of titanium dioxide (with dielectric constant 100) is inserted between the plates. Indicate whether each of the following quantities increases or decreases after insertion of the dielectric, as well as the factor by which each quantity changes: (i) potential difference across the plates; (ii) capacitance of the capacitor; and (iii) total energy stored in the capacitor. (No calculations are required here.) Calculate the total energy stored in the capacitor with dielectric described in part (b).Two parallel plate capacitors are connected to a 60 V battery as shown. C₁ is air filled and C₂ uses a dielectric material with relative permittivity/dielectric constant of -2.6. Each capacitor has a plate area of A = 80 cm² and a plate separation of 3.0 mm. a) Find the charge Q on the plates of by C₁ and C2. b) The energy stored by C₁, C2 and total energy stored. + 60V C₂ *Indicate all assumed voltage polarities C₁
- Capacitance Problem 3: Two parallel plates have charges +Q and −Q, as shown. The magnitude of the charges is 29.7nC, the area of each plate is A=0.0105 m2, and the distance between them is d = 8.53cm. The plates are surrounded by air. Part (f) If the area is doubled to 2A while the gap is decreased to 14d, then what is the value, in picofarads, of the new capacitance? Part (g) What is the new value, in volts, of potential difference between the plates with the original charges, +Q and −Q, given the increased area and reduced gap of Part (f)?answer c d eFor the system of capacitors shown in the figure below, find the following. (Let C, = 3.00 µF and C, = 1.00 µF.) %3D 6.00 µF 2.00 µF C, 90.0 V (a) the equivalent capacitance of the system µF (b) the charge on each capacitor on C1 µC on C2 on the 6.00 pF capacitor µC on the 2.00 pF capacitor (c) the potential difference across each capacitor across C, V across C, V across the 6.00 µF capacitor V across the 2.00 µF capacitor V