Continuing our discussion on Capacitors
Cylindrical Capacitors (I) Two concentric conducting cylinders of length L and radii R and R. We determine the electric field between the cylinders using Gauss s Law: we imagine a cylinder in between the other two. The enclosed charge is Q, and the field is radial: E(πrL)=Q/ε 0 R < R << L r E = V C = πlε = V Q V 0 V Q r rˆ where V is the magnitude of = R R πlε 0 = ln( R R ) r r E dr = R R Q Q dr = ln( R πlε r πlε 0 the potential difference...a 0 R ) positive quantity
Cylindrical Capacitors (II) If (R R )<< L, then the cylindrical capacitor acts much like a parallel plate capacitor. We revisit our integral for the potential difference. narrow, so we approximate the integral by taking dr V C = V V This has the form of and plate area = R R πrlε 0 R R A = r r E dr = R R πrl (the area of Q Q R dr πlε r πlε 0 The interval from R 0 the curved surface of a cylinder). ( R R ) is very / the capacitance of a parallel plate capacitor with spacing d = R R R R R and r to R R. R
Spherical Capacitors Two concentric conducting spheres of radii R and R. (hard to build ) R < R We know already the field between the two spheres: it is radial with magnitude E=kQ/r. R Q Q R V C = = V Q V = V k = R R R R r r E dr = = 4πε 0 ( R R ) R R R R Q k dr r R R = kq R R
A parallel plate capcitor has an area of A = cm = X0-4 m and a plate separation of d = mm = X0-3 m. Find its capacitance. What is the capacitance if the plate separation is increased to 3mm?
Quiz: Case Case Q Q d -Q d -Q The capacitance of the conductors.
Quiz: 50 V - - 0 V Case A Case B The capacitance of the conductors.
Quiz: If the area of the plates of a parallel-plate capacitor is doubled, the capacitance is a. not changed b. doubled c. halved d. quadrupled e. quartered
Energy Stored in a Capacitor If plates of a charged capacitor are connected together by a conductor, charge will transfer from one plate to the other until the two are uncharged. The energy stored in a capacitor may be considered as being stored in the electric field created between plates as the capacitor is charged The electric field is proportional to the charge E=Q/ε o A for two plates
Energy Stored in a Capacitor (I) uncharged capacitor Ll du = Vdq= q C dq Move charge from one plate to the other This action required work --- how much? U = Q q C dq = Q C 0 This result applies to any capacitor, regardless of its geometry. As the stored energy increases as C and the potential difference V increases. = QV = CV
Energy Stored in a Capacitor (II) We can also store (or extract) energy by moving the capacitor plates. r E d,v,c Q Q r E Q Q d,v,c The work done is E ( d d). The potential increases from V to V = d d V. The capacitance decreases from C to C = d d C.
Energy Stored in the Electric Field Let s express the energy stored in a parallel plate capacitor in terms of its internal electric field: U = CV Then the energy per unit volume is u = e ε 0 A d = ε 0E ( Ed ) = ε E ( Ad ) We call this the energy density of the electric field 0 { volume inside of capacitor This expression is in fact valid for any electric field.
Combinations of Capacitors Individual capacitors can be connected into combinations. We can calculate the equivalent capacitance of these combinations. parallel connection: the capacitors have the same potential difference placed across them series connection: the capacitors have the same charge stored in them
Important: Not every connection is series or parallel!!! Although this would not usually occur in a circuit constructed by a professional, it can happen in nature, or in amateur (!) circuits, that some capacitances are neither in series nor in parallel. Each pair is a series combination, and the two pairs are in parallel to each other. Now, none of the capacitors are in series or in parallel with any other.
Parallel Capacitor Combinations The total charge stored is Q=Q Q The potential difference is V=V a V b Thus, the equivalent capacitance is C = Q V = Q Q V = Q V Q V = C C
Series Capacitor Combinations The total charge stored is Q=C V =C V The potential difference is V=V V Thus, the equivalent capacitance is C = Q V = Q V V = Q Q Q = C C C C C = C C
The equivalent capacitance for series and parallel arrangements may be determined by considering the following relationships For capacitors arranged in Series: Thus, the equivalent capacitance is C = Q V = Q V V = Q Q Q = C C C C C = C C For capacitors arranged in Parallel: Thus, the equivalent capacitance is C = Q V = Q Q V = Q V Q V = C C So a practical problem
Find the equivalent capacitance between a and b for the combination of capacitors shown below
Find the equivalent capacitance between a and b for the combination of capacitors shown below C C C 3 The equivalent capacitance of this combination is C C C 3
What is the equivalent capacitance?
Now let s discuss capacitors with dielectrics. A dielectric is a nonconducting material such as rubber glass wax What happens to a capacitor when a dielectric is inserted between the conducting surfaces? Let s consider a parallel plate capacitor.
When a dielectric is inserted between the plates of a charged capacitor The charge on the plates remains unchanged But the potential difference as recorded by an electrostatic voltmeter is reduced from V O to V = V O κ
This means that the capacitance increases by a factor, к So when a dielectric material is inserted between the plates of a capacitor, the capacitance increases. If the dielectric completely fills the space between the plates, the capacitance increases by a dimensionless factor, к, called the dielectric constant.
Dielectrics part Consider this problem: A metal slab is inserted into the capacitor. The charges in the plates induce an equal charge density on the surfaces of the slab, so that E=0 inside the slab. This action reduces the potential difference between the two plates, and so increases the capacitance. The conducting slab is completely polarized. The electric field inside the slab is completely zero.
Dielectrics part If we stick in an insulator, there will be a partial polarization at the surface, and a partial reduction of the electric field. Since the charges in the plates are free to move, we call that the free charge density, σ f. The charges in the insulator are bound to atoms (not free to move), and so we call that the bound charge density, σ b. The full electric field is E 0 = σ f ε 0. The reduced field in the insulator is E = σ f σ b ε 0
Dielectrics part 3 An insulator placed in a capacitor in this way affects the Field. Some materials have a larger effect on the Field than others. The ratio of the full Field to the reduced Field is the dielectric constant of the material. κ = E 0 E = σ f σ f σ b We also sometimes speak of the permittivity of the material. ε = κε 0
Dielectrics in Capacitors If the dielectric completely fills the volume of the capacitor, then the potential (field) is reduced by the factor k and thus the capacitance will be increased by k. for a C C = κ = κc parallel plate capacitor this ε o 0 A d may be written as
Practical reasons for using dielectrics We saw that we could use a conductor in between the plates to increase the capacitance, but there is always the chance of the inserted conductor touching the plates! Dielectrics don t have that problem. It is easier to maintain a precise separation of the two plates if there is a material in between the two plates to hold them in place. So, even just inserting a weak dielectric (κ ~ ) has a practical function. The dielectric breakdown field of air is only 3x0 6 V/m. Much higher fields can be maintained in a capacitor if a dielectric is used in the gap instead of air.
Table of Dielectrics
What if the dielectric doesn t fill the gap? = For the basic parallel plate capacitor with an air gap, C = ε 0 A d ( ( ) 0. m ) = 8.85 pf/m =. pf 0.004 m If we fill the gap with a dielectric of κ =, C = κε 0 A d = 44. pf The partly filled gap is just two capacitors in series! We add them in the usual way. C = ε 0 A d 4 ( ) κε 0 A ( 3d 4) A = ε 0 d 4 3 κ = 35.4 pf
Energy Stored in a Capacitor with a Dielectric The increase in capacitance means an increase in energy stored, for a given potential. U = CV Then the energy per unit volume is u = e ε 0 A d = ε 0E ( Ed ) = ε E ( Ad ) 0 { volume inside of capacitor
However, with a and dielectric inserted the capacitance is C = κc and so the potential energy becomes u e 0 U = CV U = κc0v the energy per unit volume becomes = κε 0 E given by
The energy of a capacitor is raised when a diaelectric is inserted. This means that Work is done on the dielectric which draws it into the capacitor. This then implies that a force must act on the dielectric which draws it into the capacitor. This force originates from the nonuniform nature of the electric field of the capacitor near its edges
The nonuniform electric field near the edges of a parallel plate capacitor causes a dielectric to be pulled into the capacitor
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