Question 17 ( The effective area of each plate of a parallel plate capacitor is 2.1 m2. The capacitor is filled with neoprene rubber (k = 6.4). When a 6.0-V potential difference exists across the plates of the capacitor, the capacitor stores 4.0 µC of charge. a) What is the capacitance of the capacitor? (- b) Determine the plate separation of the capacitor. (
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- An air-filled parallel-plate capacitor has plates of area 2.20 cm2 separated by 2.50 mm. The capacitor is connected to a 18.0-V battery. (a) Find the value of its capacitance. pF (b) What is the charge on the capacitor? pC (c) What is the magnitude of the uniform electric field between the plates? V/mA certain parallel-plate capacitor consists of two plates, each with area 2.00 x 102cm2, and separated by a gap of 0.400 cm. (Assume air as the dielectric material.) a) What is the capacitance of this capacitor? b) If the capacitor is connected across a 500.0 V source, what charge will it hold? c) What is the magnitude of the electric field between the plates of the capacitor? (under the conditions of part b)?Regarding the Earth and a cloud layer 650 m above the Earth as the "plates" of a capacitor, calculate the capacitance if the cloud layer has an area of 1.53 km2. If an electric field of 2.0 × 106 N/C makes the air breakdown and conduct electricity (lightning), what is the maximum charge the cloud can hold? εo = 8.85 x 10−12 F/m
- The plates of a spherical capacitor have radii 41.3 mm and 44.2 mm. (a)Calculate the capacitance. (b) What must be the plate area of a parallel-plate capacitor with the same plate separation and capacitance?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…Capacitance 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.
- A parallel plate capacitor is charged to a potential of 3000 V and then isolated. Find the magnitude of the charge on the positive plate if the plates area is 0.40 m2 and the distance between the plates is 0.020 m. Group of answer choices 68 µC 54 µC 27 µC 18 µCA parallel-plate capacitor in air has a plate separation of 1.58 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 270 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion. before pC pC after (b) Determine the capacitance and potential difference after immersion. Cf = F AV = V (c) Determine the change in energy of the capacitor. nJ Need Help? Read Ita) A capacitor consists of two parallel metal plates immersed in a high-dielectric liquid, chloro- cyclohexane (e = 30). When the separation of the plates is 1mm, the capacitance is 0.06 F. Initially it is charged, at 500 V. Assume that 0 = 8.85 x 10-12 Fm-1. i) == Use this information to calculate the area of the plates, and the charge and the energy stored. ii) The plates are now pulled apart, to a gap of 1cm. Calculate the voltage it is now at, the energy now stored, and the force required to pull the plates apart.