The parallel plates in a capacitor, with a plate area of 9.00 cm2 and an air-filled separation of 4.00 mm, are charged by a 4.40 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 7.40 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates. (a) Number i Units (b) Number Units (c) Number i Units (d) Number i Units
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- A model of a red blood cell portrays the cell as a spherical capacitor, a positively charged liquid sphere of surface area A separated from the surrounding negatively charged fluid by a membrane of thickness t. Tiny electrodes introduced into the interior of the cell show a potential difference of 100 mV across the membrane. The membrane's thickness is estimated to be 96 nm and has a dielectric constant of 5.00. (a) If an average red blood cell has a mass of 1.10 ✕ 10−12 kg, estimate the volume of the cell and thus find its surface area. The density of blood is 1,100 kg/m3. (Assume the volume of blood due to components other than red blood cells is negligible.) volume m3 surface area m2 (b) Estimate the capacitance of the cell by assuming the membrane surfaces act as parallel plates. (c) Calculate the charge on the surface of the membrane. How many electronic charges does the surface charge represent?In which situation's would angular acceleration be negative? Select all that apply 1. An object is at rest and is starting to rotate clockwise 2. An object is rotating clockwise and speeding up 3. An object is rotating clockwise and speeding up 4. An object is rotating counterclockwise and slowing downA parallel-plate air-filled capacitor having area 38 cm² and plate spacing 1.4 mm is charged to a potential difference of 500 V. Find (a) the capacitance, (b) the magnitude of the charge on each plate, (c) the stored energy, (d) the electric field between the plates, (e) the energy density between the plates. (a) Number (b) Number (c) Number 24 (e) Number 12E-9 0.75 (d) Number 357142 0.056 Units Units Units Units Units pF μJ N/C or V/m J/m^3
- The plates of an air-filled parallel-plate capacitor are 3.20 mm apart, and each has an area of 3.80 cm^2. The capacitor is connected to a 6.00 V battery. Calculate: (a)the capacitance; (b) the charge stored on the plates; (c) the magnitude of the electric field between the plates (d) the stored energy.Two identical parallel-plate capacitors, each with capacitance 15.5 F, are charged to potential difference 46.0 V and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy. O Positive work is done by the agent pulling the plates apart. Negative work is done by the agent pulling the plates apart. No work is done by pulling the agent pulling the plates apart.The parallel plates in a capacitor, with a plate area of 6.70 cm2 and an air-filled separation of 2.60 mm, are charged by a 7.70 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.50 mm. Neglecting fringing, find (d) the work required to separate the plates.
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