Consider a parallel plate capacitor having plates of area 1.4 cm^2 that are sepa

Question

Consider a parallel plate capacitor having plates of area 1.4 cm^2 that are separated by 0.012 mm of neoprene rubber. You may assume the rubber has a dielectric constant x = 6.7. What is the capacitance in Mu_F? What is the capacitance of a parallel plate capacitor? How does the dielectric constant change the permittivity of the material? The permittivity in the material is simply e = xe_0. The answer provided was not correct. We have recognized the following, Your answer appears to be off by a factor of 10^n, where n is an integer value. Ensure you have represented the number in the correct units. What charge, Q, in coulombs, does it hold when 9.00 V is applied to it?

Explanation / Answer

We know that C=(Epsilon)*A/d

Epsilon: Electric permittivity of the substance separating the plates

Electric permittivity of Neoprene= Relative permittivity of Neoprene * E-permittivity of free space.

Relative permittivity of Neoprene= 6.7 (Dimensionless quantity)

E-permittivity of free space= 8.854187817* 10^(12) F/m

C=( 6.7*8.854187817* 10^(12) ) * (1.40 x10^-4) / (0.012*10^-3)= 6.92*10^-13 F

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