The dielectric in a capacitor serves two purposes. It increases the capacitance,

Question

The dielectric in a capacitor serves two purposes. It increases the capacitance, compared to an otherwise identical capacitor with an air gap, and it increases the maximum potential difference the capacitor can support. If the electric field in a material is sufficiently strong, the material will suddenly become able to conduct, creating a spark. The critical field strength, at which breakdown occurs, is 3.0 MV/m for air, but 60 MV/m for Teflon.

A. A parallel-plate capacitor consists of two square plates 18 cm on a side, spaced 0.60 mm apart with only air between them. What is the maximum energy that can be stored by the capacitor?

B. What is the maximum energy that can be stored if the plates are separated by a 0.60-mm-thick Teflon sheet?

Please show work so that I can thoroughly understand the steps.

Explanation / Answer

here,

side of plates = 18 cm = .18 m

Area of plate = .18^2 = 0.0324 m^2

distnce between the plates = 0.60 mm = 0.0006 m

CASE A:

K for air = 3 MV = 3000000 V

as Energy Stored in Capacitor is given as :
U = 0.5 * C * V^2
Where
C = capacitance = eo * A /d = ( 8.85 *10^-12 * .0324 ) / 0.0006
C = 4.779 × 10^-10 F

V = K for air * Distance between plates = 3000000 * 0.0006
V = 1800 V

Therefore,

U = 0. 5 * 4.779 × 10^-10 * 1800^2
U = 7.74*10^-4 J

the maximum energy that can be stored by the capacitor is 7.74*10^-4 J

CASE A:

K for Teflon = 6 MV = 6000000 V

as Energy Stored in Capacitor is given as :
U = 0.5 * C * V^2
Where
C = capacitance = eo * A /d = ( 8.85 *10^-12 * .0324 ) / 0.0006
C = 4.779 × 10^-10 F

V = K for air * Distance between plates
V = 6000000 * 0.0006
V = 3600 V

Therefore,

U = 0. 5 * 4.779 × 10^-10 * 3600^2
U = 3.096 *10^-3 J

the maximum energy that can be stored by the capacitor is 3.096 *10^-3 J

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