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College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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
Transcribed Image Text:11. A parallel plate capacitor is constructed with plates of area 0.028 m² and
separated by 0.55 mm. Find the magnitude of the charge on each plate of the
capacitor when the potential difference between the plates is 20.1V
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- A 500 pF parallel-plate capacitor has +40 nC charges on its plates. The plates are separated by 0.4 mm. Find: a) the area of each plate; cm? b) the potential difference between the plates; V c) the electric field between the plates; N/Carrow_forwardA parallel plate capacitor is charged up so that the upper plate carries a charge of 0.2C. In a vacuum, the potential difference between the plates is 2V. When a certain dielectric material is placed between the plates, the potential difference drops to 1.5V. Calculate the dielectric constant of the material.arrow_forwardAn air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm², separated by a distance of 1.50 mm. A 25.0- V potential difference is applied to these plates. (a) Calculate the electric field between the plates. |kV/m (b) Calculate the surface charge density. nC/m² (c) Calculate the capacitance. pF (d) Calculate the charge on each plate. pCarrow_forward
- Problem 5 Consider a parallel-plate capacitor with a plate area of A = 8.50 cm². The separation between the plates is d₂ = 3.00 mm (the space between the plates is filled with air). The plates of the capacitor are charged by a 6.00 V battery, i.e., the potential difference between the plates is V₂ = 6.00 V. The plates are then disconnected from the battery and pulled apart (without discharge) to a sepa- ration of df = 8.00 mm. In the following, neglecting any fringing effects. (a) Will the new potential difference between the plates be larger, smaller, or the same compared to the initial potential difference of V₂ = 6.00 V? Explain. (Hint: Note that the charge will not change when the plates are pulled apart. Why is that?) (b) Find the potential difference Vf between the plates after the plates have been pulled to their new, larger separation df. (c) Find the electrostatic energy stored in the capacitor before and after the plates are pulled apart. (d) To separate the plates, you will…arrow_forwardCapacitance Problem 18: A cylindrical capacitor is made of two concentric conducting cylinders. The inner cylinder has radius R1 = 19 cm and carries a uniform charge per unit length of λ = 30 μC/m. The outer cylinder has radius R2 = 45 cm and carries an equal but opposite charge distribution as the inner cylinder. Part (b) Calculate the electric potential difference between the outside and the inside cylinders in V. Part (c) Calculate the capacitance per unit length of these concentric cylinders in F/m.arrow_forward1) solve this question, as soon as possible.arrow_forward
- A parallel-plate capacitor filled with air has plates of area 6.5×10−3 m2 and a separation of 0.32 mm. a) Find the magnitude of the charge on each plate when the capacitor is connected to a 11-V battery. (Q = ? nC) b) Why does answer A decrease if the seperation between the plates increases? c) Calculate the magnitude of the charge on the plates if the separation is 0.90 mm. (Q = ? nC)arrow_forwardWhen a potential difference of 132 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 20.0 nC/cm². What is the spacing between the plates? μmarrow_forward
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