Question

A parallel-plate capacitor has plates of area 3.91×10−4 m2 .

A) What plate separation is required if the capacitance is to be 1510 pF ? Assume that the space between the plates is filled with air? (Dielectric constant for air is 1.00059)
B) What plate separation is required if the capacitance is to be 1510 pF ? Assume that the space between the plates is filled with paper. (Dielectric constant for paper is 3.7)

Answer 1

To find the plate separation of a parallel-plate capacitor, use the formula d = (ε*A)/(C*η). For part A, with air as the dielectric, the plate separation is approximately 0.618 mm. For part B, with paper as the dielectric, the plate separation is approximately 4.79 mm.

Step-by-step explanation:

To calculate the plate separation of a parallel-plate capacitor, we can use the formula:



d = (ε*A)/(C*η)



where d is the plate separation, ε is the permittivity of the dielectric material, A is the area of the plates, C is the capacitance, and η is the number of plates. For part A, since the space between the plates is filled with air and the dielectric constant for air is approximately 1.00059, we substitute this value into the formula and calculate d:



d = (1.00059*3.91e-4)/(1510e-12*2) = 0.000618 m = 0.618 mm



For part B, since the space between the plates is filled with paper and the dielectric constant for paper is 3.7, we substitute this value into the formula and calculate d:



d = (3.7*3.91e-4)/(1510e-12*2) = 0.00479 m = 4.79 mm