Anonymous
Anonymous asked in Science & MathematicsEngineering · 9 years ago

# An air-filled parallel-plate capacitor has a capacitance of 1.6 pF.......?

An air-filled parallel-plate capacitor has a capacitance of 1.6 pF. The separation of the plates is doubled and wax is inserted between them. The new capacitance is 2.8 pF. Find the dielectric constant of the wax.

Relevance

Please study this matter, Capacitance and Capacitors

I. Elementary Characteristics

In its most elementary state a capacitor

consists of two metal plates separated by

a certain distance d, in between the plates

lies a dielectric material with dielectric

constant έ = εoε , where εo is the dielectric of air.

The dielectric material allows for charge to accumulate between the capacitor plates. Air (actually vacuum) has the lowest dielectric value of εo = 8.854 * 10-12 Farads/meter where the Farad is the unit for capacitance.

All other materials have higher dielectric values, since they are higher in density and can therefore accumulate more charge.

C = εoε * A/ d,

In your question C1/C2 = ε1* d2/ε2*d1, or 1.6/2.8 = 8.854*10^-12 * 2/ ε2

accordingly calculate the value of ε2

• C = 1.6 pF = (epsilon*Area*k)/d

rearranging,

d = (epsilon*Area*k)/1.6pF

where,

k=1.0

epsilon = 8.854*10^-12 F/m

area = unknown

doubling the separation,

d = (2*area*1.0*8.854*10^-12)/(1.6*10^-12)

d = 11.0675Area

Now that d will be plugged into the new equation

2.8 pF = (k*area*epsilon)/d

2.8 pF = (k*area*epsilon)/(11.0675Area)

areas cancel out, and you solve for k

k = 3.5