The capacitance C of parallel plate capacitors with area and are separated by a distance of air as follows.
Here, is the permittivity of free space.
Convert the distance between the plates into meters.
Calculate the area of the plates by using the above equation.
Substitute for, for and for.
Therefore, the area of the plates is.
A 0.20-F capacitor is desired . What area must the plates have if they are to...
A 0.31 F capacitor is desired. What area must the plates have if they are to be separated by a 3.2 mm air gap? In m2
0.31 F capacitor is desired. What area must the plates have if they are to be separated by a 3.2 mm air gap? In meters squared
A 0.17-F capacitor is desired. What area must the plates have if they are to be separated by a 3.2-mm air gap?
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Capacitor calculation a) Calculate the capacitance of a parallel-plate capacitor whose plates are made of two different sizes. One plate has a radius of 10 cm and the other plate has 12 cm and is separated by 0.75 mm air gap. b) What is the charge on each plate if a 12-V battery is connected across the two plates? c) What is the electric field between the plates? d) Estimate the area of the plates needed to achieve a capacitance...
6) Capacitor calculation a) Calculate the capacitance of a parallel-plate capacitor whose plates are made of two different sizes. One plate has a radius of 10 cm and the other plate has 12 cm and is separated by 0.75 mm air gap. b) What is the charge on each plate if a 12-V battery is connected across the two plates? c) What is the electric field between the plates? d) Estimate the area of the plates needed to achieve a...
An electric field of 9.00×105 V/m is desired between two parallel plates, each of area 45.0 cm2 and separated by 2.45 mm of air. What charge must be on each plate?
An electric field of 9.50×105 V/m is desired between two parallel plates, each of area 45.0 cm2 and separated by 2.45 mm of air. What charge must be on each plate?
A parallel plate capacitor is made of plates of area 0.05 m? each. The plates are separated by a distance of 0.200 mm. Initially, the space between the plates is filled with air. (a) What is the capacitance of this air-filled capacitor? (b) If the electric field inside the capacitor exceeds 3.00 x 106 V/m, the air undergoes electrical break- down. (This maximum field is known as the dielectric strength of air.) From this, calculate the maxi- mum voltage (potential...
Area of a Capacitor FR If a capacitor has 4.64 μC of charge on it and an electric field of 2.41 kV/mm is desired if they are separated by 3.93 mm of air, what must each plate's area be? Submit Answer Tries 0/10