Capacitance and Dielectrics ------------------------------------ Joseph F. Alward, PhD     Department of  Physics   University of the Pacific

   The Parallel-Plate Capacitor                            

Equal and opposite charges are placed on the plates of parallel metal plates. Electric field E points from positive plate to the negative plate.

  Capacitors and Dielectrics        

A dielectric is any polarizable material.  All materials are polarizable, so all materials are dielectrics, even air.

  Defining Capacitance                                                         

C = "capacitance"     = q /DV Units:  coulomb/volt            = farad (F) ----------------------------- The capacitance of a capacitor is constant; if q increases, DV increases proportion- ately.      Michael Faraday          (1791-1867) --------------------------------------------------------- Most authors omit the D; thus,       C = q /V

   Capacitance and Heat Capacity         

Compare this to  DV= Q/C  for charge Q and  capacitance C

The greater the heat capacity C of an object, the smaller will be its temperature rise   DT for a given heat input Q.   Thus, DT = Q/C If an object has a large heat capacity C, it will take a lot of heat Q increase its temperature, or to melt it. C = Q/DT = constant

  Exceeding the Capacity of the Capacitor                   

The greater the capacitance of a pair of conductors, the smaller will be the rise in potential,   DV, for a given addition of charge,   Dq, to each conductor. Thus, DV =   Dq/C If a capacitor has a large capacitance C, a large charge   Dq may be added  to the conductors without causing a large increase,  DV, in the potential difference.

   Examples of Capacitors                                                         

  Calculating Capacitance

Problem:  What is C?

Solution: q = 20 nC V = 10 V C =q/V    = 20 x 10-9/10    = 2.0 x 10-9 F    = 2.0 nF

   Calculating the Charge Stored on a Capacitor          

Problem:  A 4 mF capacitor is connected to a 12-volt battery. What charge q is stored on each side of the capacitor ? ----------------------------------------------------------- Solution:  q = CV    = (4 x 10-6) 12    = 48 x 10-6 coulomb    = 48 mC ----------------------------------------------------------- What will happen to the charge on the capacitor when the switch is opened?

  Energy Stored in a Capacitor            

Three ways to calculate the same thing. U = (1/2) Q2/C     (1) -------------------------------- Use C =Q/V to obtain   U=  (1/2)QV        (2) -------------------------------- and U = (1/2) CV2      (3)

    Energy Stored in a Capacitor                                           

Problem:  The capacitor in the figure has a capacitance of 10 x 104 pF.  It is charged to a potential V = 300 volts. How much energy is stored in the capacitor? ------------------------------------------------------------- Solution:  One picofarad is 1 x 10-12 F. 10  x 104 pF = 10 x 104 x 10-12 F                     = 10 x 10-8 F U = (1/2)CV2     = (1/2)(10 x 10-8)(300)2     = 0.003 J

  Capacitance of a Parallel-Plate Capacitor                  

C = Ae0/d   where e0 = 8.85 x 10-12 (SI units) ---------------------------------------------------------------- Example:  Two square metal plates are 20 cm on a side and separated by one millimeter.   What is their capacitance? C = (0.20)2 (8.85 x 10-12)/0.001     = 3.54 x 10-10 F     = 354 pF ---------------------------------------------------------------- Person-Ground capacitance:  100 pF Cow-Ground capacitance:  200 pF

  Effect of Dielectric on the Capacitance                             

Parallel-plate capacitor with an air dielectric.  If E exceeds 30,000 volts/cm, arc discharge (lightning) will occur.

Layer of charge is induced on surface of dielectric

Effective charge on each plate is reduced, lowering the potential difference DV and the electric field E.  This allows more charge to be put on plates; the "capacity" to carry charge is increased.

   Dielectrics Reduce E and Increase C                                

Dielectric constant = k            (kappa) New potential difference: DV =   DV0/k New electric field E: E = E0 /k -----------------------  Old capacitance:  C0 = Q0/DV0  New capacitance:  C = Q0/(DV0/k)      = k Q0/DV0      = k C0

Dielectric Constants             Substance    k Vacuum 1 Air 1.00054 Paper (royal grey) 3.3 Ruby mica 5.4

A capacitor filled with ruby mica will store 5.4 times as much energy  as an air-fillled capacitor.

 

Example:  If the electric field E between the plates of a capacitor is 500 V/m in air, then it   will only be 500/5.4 = 92.6 V/m if ruby mica completely fills the region between the plates.

  Applications                

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