
College Physics
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ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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An air-filled parallel-plate capacitor has a capacitance of 1.2 pF. The separation of the plates is doubled and wax is inserted between them. The new capacitance is 3.1 pF. Find the dielectric constant of the wax
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- 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?arrow_forwardA certain substance has a dielectric constant of 3.8 and a dielectric strength of 25 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 5.3 × 10-2 µ F and to ensure that the capacitor will be able to withstand a potential difference of 3.1 kV?arrow_forwardA 1.00-?F capacitor is charged by being connected across a 9.0-V battery. It is then disconnected from the battery and connected across an uncharged 8.00-?F capacitor. Determine the resulting charge on each capacitor.arrow_forward
- The capacitor is charged to a voltage of 4.00 kv using a power source that is then removed. The gap between the plates is then filled by a dielectric layer. The charge on each plate stays constant at 2.50 kv despite the reduction in the potential difference between the plates. Calculate the initial capacitance value of the system.arrow_forwardAn air-filled parallel-plate capacitor has a capacitance of 2 pF. The plate separation is then halved and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 8 pF. The dielectric constant of the wax is:arrow_forwardConsider a parallel plate capacitor having plates of area 1.4 cm2 that are separated by 0.018 mm of neoprene rubber. You may assume the rubber has a dielectric constant κ = 6.7. a) What is the capacitance in μF? b) What charge, Q, in coulombs, does it hold when 9.00 V is applied to it?arrow_forward
- A parallel plate capacitor with a capacitance of 0.86 F with a has plate area A and are separated by a distance d. It is charged up to 11 V. The capacitor is now disconnected from the battery. If the area of the plates increases by a factor of 3 and the separation decreases by a factor of 10, find the new charge on the capacitor in Coulombs.arrow_forwardAn air-filled capacitor is made from two flat parallel plates 1.2 mm apart. The inside area of each plate is 8 cm². (a) What is the capacitance of this set of plates? C F (b) If the region between the plates is filled with a material whose dielectric constant is 6.5, what is the new capacitance? C dielectric air = = Farrow_forward
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