What capacitance is needed to store 2.50 mu C of charge at a voltage of 220 V? F

Question

What capacitance is needed to store 2.50 mu C of charge at a voltage of 220 V? Find the capacitance of a parallel plate capacitor having plates of area 4.00 m^2 that are separated by 0.200 mm of Teflon. A large capacitance of 1.20 mF is needed for a certain application. Calculate the area the parallel plates of such a capacitor must have if they are separated by 3.50 pm of Teflon. What is the maximum voltage that can be applied? Find the maximum charge that can be stored. Calculate the volume of Teflon alone in the capacitor. Find the total capacitance of the combination of capacitors shown in Figure 18.24. (C_1 = 20.0 mu F, C2 = 2.00 mu F.)

Explanation / Answer

here,

1)

charge , q = 2.5 * 10^-6 C

V = 220 V

cpacitance , C = q/V

C = 1.14 * 10^-8 F

the capacitance is 1.14 * 10^-8 F

2)

area = 4 m^2

sepration between plates , d = 0.002 m

capacitance , C = area * e0 /d

C = 4 * 8.85 * 10^-12 /0.002

C = 1.77 * 10^-8 F

C = 0.0177 uF

3)

capacitance , C = 1.2 * 10^-3 F

(a)

sepration between the paltes , d = 3.5 * 10^-6 m

let the area of plates be s

C= a * e0 /d

0.0012 = a * 8.85 * 10^-12 /( 3.5 * 10^-6)

a = 474.57 m^2

the area of plates is 474.57 m^2

(b)

You should know the value for the critical electric field Ec in air (the electric field that would cause electric breakdown) and calculate the voltage from the formula:

E = V / d

Ec = 3 * 10^6 V/m

maximum voltage , Vm = Ec * d

Vm = 3 * 10^6 * 3.5 * 10^-6

Vm = 10.5 V

the maximum voltage that can be applied is 10.5 V

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