find the energy stored with a dielectric (Er = 8.5) is placed between the plates with the gap between them is decreased by 15% and the voltage goes up to 20 kV.
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If you have a capacitor that stores 2.5 J of energy with an aur gap and power supply at 10 kV, find the energy stored with a dielectric (Er = 8.5) is placed between the plates with the gap between them is decreased by 15% and the voltage goes up to 20 kV.
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- Shown in the figure below is a capacitor. Each plate has an area of A = 3.94e-02 m2 and has a gap d = 5.1e-04 meters. At the first moment, the capacitor is empty. At the second moment is it partly filled with a dielectric of constant K = 2.64. The thickness of the dielectric layer is a = 2.04e-04 meters leaving a free space of b = 3.06e-04 meters.Write the FORMULA for the capacitance of the empty capacitor. Determine the capacitance of the empty capacitor Cempty = FaradsWhen you partly fill the capacitor, which of these best describes the new configuration: Neither series nor parallelSeries ParallelBoth series and parallel Determine the formula for the partly filled capacitor Cfilled = Determine the value of the partly filled capacitor Cfilled = FaradsA capacitor is constructed from two conducting disks with radii of r = 2.0 cm and adielectric with constant κ = 270.(a) At what separation distance d should the plates be to give a capacitance of 3.0 nF?Assume the dielectric can be made to fill the space completely for any d. (b) If this capacitor is connected to a battery with voltage E = 12.0 V, how muchcharge Q will collect on the plates? (c) How much electric potential energy UC will be stored in the charged capacitor asa result?A parallel plate capacitor has plates of area A = 5.50 x 10 m² separated by distance d = 2.37 × 10-4 m. (The permittivity of free space is & = 8.85 × 10¯ -2 -12 HINT (a) Calculate the capacitance (in F) if the space between the plates is filled with air. 2.05e-9 F What is the capacitance (in F) if the space is filled half with air and half with a dielectric of constant x = 2.30 as in figure (a), and figure (b)? (Hint: One of the capacitors is a parallel combination and the other is a series combination.) (b) figure (a) 2.85e-9 (c) figure (b) a F 1.90e-16 X F A K b A c²/(N • m²).) K
- An air capacitor has a capacitance of 2 µF, which becomes 12 µF when a dielectric medium is filled in the space between the plates. Find (i) dielectric constant of that material.A certain capacitor has a capacitance of 4μF when its plates are separated by 0.2 mm of vacant space. A battery is used to charge the plate to a potential difference of 500 V and is then disconnected from the system. (a) What will be the potential difference across the plates if a sheet of mica 0.2 mm thick is inserted between the plates? (b) What will the capacitance be after the dielectric is inserted? (c) What is the permittivity of mica? NOTE: My professor said that the permitttivity of mica is 4.8 x 10^-11 C2/Nm2Find the capacitance of a parallel-plate capacitor having plates with a surface area of 7.49*10-2 m2 and separated by 3.01 mm of Neoprene (which has a dielectric constant of 6.7). Answer in nanoFarads.
- (a) When a 6.50-v battery is connected to the plates of a capacitor, it stores a charge of 32.0 µc. What is the value of the capacitance? (b) If the same capacitor is connected to a 17.5-V battery, what charge is stored?A 72-nF parallel-plate capacitor has platesthat sandwich a dielectric layer that is 75 mm thick. (a) In orderto reduce the capacitance, should the thickness of the dielectric layer be increased or decreased? (b) Calculate the thickness of thedielectric layer that would reduce the capacitance to 64 nF.MY YSY When an air-filled parallel-plate capacitor is connected to a 28 V battery, it contains 80 µC of charge. After disconnecting the capacitor from the battery, the distance between two parallel plates is now reduced to half. (a) What is its new capacitance? Cnew µF %3D (b) What is the new potential difference between two plates? Vnew volts (c) Find the energy stored in the new capacitor. Unew %3D
- A large capacitance of 1.15 mF is needed for a certain application.(a) Calculate the area the parallel plates of such a capacitor must have if they are separated by 4.98 µm of Teflon, which has a dielectric constant of 2.1.m^2 (b) What is the maximum voltage that can be applied if the dielectric strength for Teflon is60 ✕ 106 V/m V (c) Find the maximum charge that can be stored.C (d) Calculate the volume of Teflon alone in the capacitor.m^3A certain capacitor has a capacitance of 4μF when its plates are separated by 0.2 mm of vacant space. A battery is used to charge the plate to a potential difference of 500 V and is then disconnected from the system. (a) What will be the potential difference across the plates if a sheet of mica 0.2 mm thick is inserted between the plates? (b) What will the capacitance be after the dielectric is inserted? (c) What is the permittivity of mica?Shown in the figure below is a capacitor. Each plate has an area of A = 3.45e-02 m² and has a gap d = 6.7e-04 meters. At the first moment, the capacitor is empty. At the second moment is it partly filled with a dielectric of constant K = 2.57. The thickness of the dielectric layer is a = 2.01e-04 meters leaving a free space of b = 4.69e-04 meters. d Empty Capacitor Area = A a b Partly Filled Capacitor Write the FORMULA for the capacitance of the empty capacitor. Determine the capacitance of the empty capacitor Cempty = Determine the formula for the partly filled capacitor Cfilled = Determine the value of the partly filled capacitor Cfilled = & A d When you partly fill the capacitor, which of these best describes the new configuration: O Both series and parallel O Neither series nor parallel Series O Parallel dielectric constant K Farads Farads NOTE: Decide whether the partly filled capacitor acts like parallel or series by following the charge: • If the charge splits between the caps,…