Thursday, February 18, 1999
Announcements: No recitation quiz next week
                          Lab Quiz next week on the concepts covered in labs 3 and 4
                          Recitation quiz for February 26 or March 1 will cover chapter 25.

Lecture notes:

• A capacitor is a device to store charges ( and hence store electrical energy). Capacitance is a measure of how effective a capacitor is.
• Let's consider of simple circuit: 
• A battery is connected to a pair a parallel plates.
• This can be shown as the following diagram:

• A battery is a device  that through chemical reactions maintains a potential difference (V) between its positive and negative terminals.
• When the battery is connected to the pair of parallel conducting plates, ( some of the) electrons in the red plate are driven to the positive terminal and an equal amount of electrons are driven from the negative terminal to the black plate.
• The set of parallel plates acts as a capacitor.
• The amount of charge the plate can hold is proportional to the potential applied ( by the battery) on the plate
• Q = CV
• Q is charge measured in coulombs
• V is potential measured in volts
• C is Capacitance and its unites are coulomb/ volt
•  A coulomb/ volt is equal to one Farad
• On Farad is a huge amount; we will usually deal with mF ( 10^-6 F) or pF ( 10^-12)

• The capacitance depends on the physical characteristics of the capacitor (i.e. the geometry)
• Parallel plate capacitor:

• The potential (or voltage) difference between the positive and negative plates is V.
• We know that E = s /e_o = Q / Ae_o
• V = - I (integral) E (dot product) ds = Ed = Qd/ Ae_o
• Q = CV or V= Q/C ; therefore C = Ae_o /d
• The larger the area of the plate, the more charge the plate can hold
• As d gets smaller, the capacitance increases
• Cylindrical Capacitor

• What is the electric field in the gap?
• I (Integral)  E (dot product) dA =  Q_enclosed / e_o
• E ( 2pr L) = Q_enclosed / e_o
•  E = Q_enclosed /( 2prL e_o )
• What is the potential in the gap?
• V_b -V_a = - I (Integral from a to b)  E (dot product) ds = - I (Integral from a to b)  Er dr
=  - [Q_enclosed /( 2prL e_o )] I (Integral from a to b) dr/r = - [Q_enclosed /( 2prL e_o )] ln (b/a)
•  C= Q/V ; therefore C = ( 2prL e_o )/ ln (b/a)
• To make the capacitance very large, decrease the distance between a and b

• Spherical capacitor

• What is the electric field in the gap?
• I (Integral)  E (dot product) dA =  Q_enclosed / e_o
• E  = Q_enclosed / ( 4pr² e_o )
• What is the potential in the gap?
• V_b -V_a = - I (Integral from a to b)  E (dot product) ds = - I (Integral from a to b)  Er dr
=  - [Q_enclosed /(  4pe_o  )] I (Integral from a to b) dr/r² = + [Q_enclosed /(  4pe_o  )] [ 1/r ] evaluated from a to b
= [Q /(  4pe_o  )] [ (1/b) - (1/a) ] = [Q /(  4pe_o  )] [ (a-b)/(ab)]
•  So if V = V_a - V_b =  [Q /(  4pe_o  )] [ (b-a)/(ab)]
• Then C= Q/V = (  4pe_o  )/ [ (ab)/(b-a)]
• To make capacitance large, a-b needs to be very small

• If we let b go to infinity, what happens?
•  C =  (  4pe_o  )/ [ (ab)/(b-a)] goes to C =  (  4pe_o  )/ (a)
• Therefore the Capacitance of an isolated sphere of radius a is  (  4pe_o  )/ (a)

• The potential difference across each of the three capacitors is the same; it is V and it is maintained by the battery
• The changes in each capacitor are:
• q₁ = C₁V
• q₂ = C₂V
• q₃ = C₃V
• The total charge on all three capacitors is :
• q_total  = q₁ + q₂ + q₃ = ( C₁+ C₂+ C₃)V
• This process is the some for any number of capacitors in parallel

B) Capacitor in series

• V = voltage through C1 (V1) + voltage through C2 (V2)
• When the battery is connected, some electrons from the top plate of C1 are transferred to the bottom plate of C2 (through the battery). As the negative charges accumulate at the bottom plate of C2, an equivalent amount of negative charges are driven away from the top plate of C2 to the bottom plate of C1.
• Summary: The amount of charges "stored" in both C1 and C2 are equal to Q
• Notice that the amount of charges "transferred" by the battery is not the sum of the charges on C1 and C2, but only Q.
• For an equivalent circuit, C = Q/V
• V = V1 + V2
•  V1 = Q/ C1 and V2 = Q/ C2
• Therefore, V = (Q/C1) + (Q/C2) = (Q/ C_equivalent)
• Hence, (1/ C_equivalent) = (1/ C1) + (1/C2)
• The process is the same if there are more capacitors in series
• In general, when capacitors are in parallel, the capacitance increases and when the capacitor are in parallel, the capacitance decreases.