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.
- Ramakant SLv 59 years agoBest Answer
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
- JacobLv 59 years ago
C = 1.6 pF = (epsilon*Area*k)/d
d = (epsilon*Area*k)/1.6pF
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