PhysicsLAB Resource Lesson
Dielectrics: Beyond the Fundamentals

Gauss' Law has shown us that the electric field between two parallel plates
 
 
can be calculated with the formula
 
 
 
If the strength of the electric field between the plates becomes too strong, then the air between them can no longer insulate the charges from sparking, or discharging, between the plates. For air, this breakdown occurs when the electric field is greater than 3 x 106 V/m. In order to keep this from happening, an insulator, or dielectric, is often inserted between the plates to reduce the strength of the electric field, without having to reduce the voltage being placed across the plates.
 
A dielectric is a polar material whose electric field aligns to oppose the original electric field already established between the plates. The dielectric is measured in terms of a dimensionless constant, κ ≥ 1, whose value is usually referenced from a table.
 
Material   κ
air   1.00054
ethanol   24.3
glass   5-10
mica   3-6
paper   2-4
paraffin   2.1-2.5
polystyrene   2.3-2.6
porcelain   5.7
rubber   2-3
teflon   2.1
water   80
 
If this insulating material is insufficient then the capacitor can still leak allowing current to flow between the plates. When this occurs the electric device "smells as if something is burning."
 


 
κ = Eoriginal / Edielectric
κ = Eo / Ed

κ = Cdielectric / Coriginal
κ = Cd / Co
 
When the battery is removed, the dielectric will decrease the electric field strength and the voltage between the plates while it increases their capacitance.
 
E = V/d
 
Using the fact that V = Ed and that capacitance is the ratio of the charge stored per unit volt we derived the following formula for the capacitance based on the geometry of a parallel-plate capacitor.
 
 
Refer to the following information for the next five questions.

A 90 µF capacitor is initially charged to 12 volts without a dielectric in place.
 If the battery remains connected while a paper dielectric having a constant of κ = 2.0 is inserted between the capacitor's plates, then what will be its new capacitance?

 How much charge will be on its plates?

 If instead, the capacitor is disconnected from the battery before the paper dielectric is inserted, what would be the capacitor's new capacitance?

 How much charge would be on its plates?

 What would be the new voltage across its plates?

For more practice with capacitors and both connected and disconnected batteries, reference this worksheet.
 
 
Dielectric Configurations
 
Often capacitors can have complicated dielectric configurations based on available materials and circuit requirements. The good news is that they can be solved based on the principles of capacitors in series and in parallel, in combination with the formulas for the electric field and the geometry of a parallel plate capacitor.
 
If the capacitors are arranged in series (one after another along a single path), then
 
Cseries = (1/C1 + 1/C2 + 1/C3)-1
 
If the capacitors arranged in parallel (strung along multiple paths that cross the same section), then
 
Cparallel = C1 + C2 + C3
 
Let's work a few examples.
 
Refer to the following information for the next seven questions.

In each of the following cases, the charged plates are 10 cm by 20 cm and the gap between the plates is 6 mm.
 
  • cream signifies a gap only filled with air, κ = 1.0
  • orange represents a dielectric with κ = 2.0
  • yellow represents a dielectric with κ = 3.0
  • purple represents a dielectric with κ = 4.0
 
What is the capacitance of this air-filled capacitor?
 
 

 
If a metal conducting slab 2 mm thick is placed in the air gap between the plates, what would be the capacitor's new capacitance?
 
 

 
What is the new capacitance if the gap is completely filled with a dielectric having κ = 3.0?
 
 

 
What is the capacitance when its gap is only half-filled with a dielectric having κ = 3.0?
 
 

 
What is the capacitance when its gap is half-filled with a dielectric having κ = 3.0 and the other half is filled with a dielectric having κ = 2.0?
 
 

 
How is the capacitance changed when the gap is half-filled with a dielectric having κ = 3.0 and the other half is filled with a dielectric having κ = 2.0 but in this new orientation?
 
 

 
And what is the capacitance of this final configuration?
 
 





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