A) A certain substance has a dielectric constant of 2.8 and a dielectric strengt

Question

A)  A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18.0 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 6.00×10-8 F and to ensure that the capacitor will be able to withstand a potential difference of 5.4 kV?

B)  A Nichrome heater dissipates 480 W when the applied potential difference is 110 V and the wire temperature is 725°C. What would be the dissipation rate if the wire temperature were held at 200°C by immersing the wire in a bath of cooking oil? The applied potential difference remains the same, and for Nichrome at 725°C is 4.00×10-4/K.

Explanation / Answer

A)

given data

capacity of capacitor, C = 6 x 10^-8 F

dielectric constant, k = 2.8

formula for the capacitance of a capacitor with dielectric material placed between

the plates is given by,

C = k0A/d

A/d = C/k0

= (6 x 10^-8 )/2.8*(8.87 * 10^-12)

=2415 m

d = A / 2415m

Dielectric strength = 18 M V/m = 18 * 10^6 V/m

(18 M V/m)A = potential difference = 5.4 kV

(18 M V/m)*A/2415 = 5400

A =5400* 2415 / 18 * 10^6

=0.724 m^2

the  minimum area should the plates of the capacitor is = 0.724 m^2

B)

Resistance at 725C is:
P = E²/R
480 = 110²/R
R =25.20 ohms

Rt = Ro(1 + alpha T)
Rt = 25.2 (1 + 4*10^-4*(–525))
Rt = 19.4ohms
assuming that the temperature coefficient for Nichrome is positive.
P = E²/R
P = 110²/19.4 = 623.7watts

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