Lesson 11 (1) Connection of Capacitors in Parallel In a parallel connection of capacitors of capacitances C1
and C2 as shown, the potential differences of the two are the same: V1 = V2 = V while the total charge stored is the sum: Q1 + Q2 = Q Substituting the following relations into the above Q1 = C1V1 = C1V Q2 = C2V2 = C2V we find (C1 + C2 ) V = Q Therefore the combination behaves as a single capacitor with the equivalent capacitance Q
C = = C1 + C2 V
The relation can be generalized to arbitrary number of capacitors in parallel: C = C1 + C2 + C3 +! (2) Capacitors in Series The following diagram shows two capacitors connected in series: The two are initially uncharged, so that the connecting wire connects plates with opposite charges of the same magnitude Q , which can be considered the stored charge in the combination. Therefore Q1 = Q2 = Q 1 On the other hand, the total potential difference from the left most to the right post plates is the sum V1 +V2 = V Substituting Q Q
Q
Q
V1 = 1 =
V2 = 2 =
C1 C1
C2 C2 !1 1$
we find # + & Q = V " C1 C2 %
The combination behaves as a single capacitor with equivalent capacitance C satisfying 1 V 1 1
= = +
C Q C1 C2
In general, for series combination of an arbitrary number of capacitors, 1 1 1
1
= + + +! C C1 C2 C3
For only two capacitors, we can use CC
C = 1 2 C1 + C2
Some networks can be decomposed into series and parallel combinations as the following example shows. Example: In the network shown, C1 = 6F, C2 = 3F C3 = 12F . A potential difference V = 8.0V exists between the terminals. Find the charges and potential differences on each capacitor. 2 Solution: First find equivalent capacitance. C12 = C1 + C2 = 6 + 3 = 9 CC
12 ! 9
C= 1 2 =
= 5.14F C1 + C2 12 + 9
Then find total charge: Q = CV = 5.14 ! 8 = 41.1C Then find individual charges and voltages: Q3 = Q = 41.1C Q 41.1
V3 = 3 =
= 3.43V C3 12
V1 = V2 = V12 = V !V3 = 8 ! 3.43 = 4.57V Q1 = C1V1 = 6 ! 4.57 = 27.4C
Q2 = C2V2 = 3! 4.57 = 13.7C (3) Reconnection of capacitors Suppose C1 , C2 carrying charges Q1,Q2 respectively are connected so the positive plates are connected together as do the negative plates. In the diagram, the positive plates are denoted by a, c and the negative by b, d .
3 The capacitors are clearly in parallel, with total charge Q1! + Q2! = Q1 + Q2 , Since the equivalent capacitance is !! +!! , the final potential difference for both capacitors is Q + Q2
V ! = Vac! " Vbd! = 1
C1 + C2
On the other hand, if the charged capacitors are connected so that the positive plate of one is connected to the negative plate of the other as shown, the capacitors are again in parallel, but the total charge is Q1! + Q2! = Q1 " Q2 . The final potential difference is V ! = Vad! " Vbc! =
Q1 " Q2
C1 + C2
(4) Dielectric Constant When a slab of dielectric material is placed in a region where there is a uniform electric field of strength E0 , induced charges appear on the surfaces of the material due to polarization. The induced charge produces an induced electric field E ind in the opposite direction. As a result, the electric field strength inside the material is reduced, being given by ! = !! − !!"# . The dielectric constant is defined by 4 !=
!!
!
and is a property of the material greater than one. The dielectric strength of the material is the electric field inside the material when it breaks down. (becoming conducting) (5) Dielectric in Capacitors Consider a parallel plate capacitor of plate area ! and plate separation ! with a dielectric filling the space between the plates. When charge !is stored in the capacitor, it resides on the plates, but not the dielectric. This charge produces the electric field !! in the absence of the dielectric given by !
!! =
!! !
The actual electric field inside the dielectric is !!
!
!=
=
!
!"! !
The potential difference between the plates is !"
! = !" =
!"! !
The capacitance is !
!
! = = !!! !
!
Thus, capacitance can be increased by the use of dielectrics. Let the induced charge on the face of the dielectric in contact with the positive plate be denoted by −!!"# . Together with the charge !on the positive plate, the charges behave like a sheet charge ! − !!"# on an area !, with the opposite charge on the negative plate. The interior electric field is related to these sheet charges by ! − !!"#
!=
!! !
5 Using the relation between ! and !! , we find Q ! Qind
Q
=
!0 A
"! 0 A
Solving for the induced charge: !−1
!!"# =
! !
Example: Find the capacitance of a parallel plate capacitor half filled with a dielectric in the space between the plates. Solution: With charge ! on the plate, the electric field is !
!" !ℎ! !"!""# !"#$%& !! !
!=
!
!" !ℎ! !"#$#%&'"% !"! !
The potential difference is !=
! !
! !
+
!! ! 2 !"! ! 2
!=
!
2!
!
=
!! ! !+1 !
The capacitance is 6