General Physics II. Conducting concentric spheres Two concentric spheres of radii R and r. The potential difference between the spheres is

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1 apacitors and Dielectrics The ideas of energy storage in E-fields can be carried a step further by understanding the concept of "apacitance" onsider a sphere with a total charge, Q, and a radius, R From previous problems we know Q that the potential at the surface is, = k R Putting more charge on the sphere stores more energy, but the ratio of energy or potential to charge depends only on R, not on Q or That is, Q R = = 4π 0R k It's true for all charged objects that the ratio of potential to voltage depends only on the shape, so this ratio is defined as the capacitance Q The units of capacitance are 1 coulomb 1 Farad 1 F ommon values of capacitance are volts microfarads, µf (10-6 Farads) and picofarads, pf (10-1 Farads) onsider two conductors connected to the terminals of a battery The battery will supply an ual amount of charge, but of opposite sign, to each of the conductors The question arising at this point: what will be capacitance of the conductor system? Let us consider different conductor systems: Parallel plates Two conducting parallel plates separated by a distance d with charges +Q and -Q The potential difference between the plates (from one plate to the other) is ρ s Qd a b = = Ed = d = o o The capacitance is Q = = d 0 onducting concentric spheres Two concentric spheres of radii R and r The potential difference between the spheres is Q 1 1 a b = = 4π o R r The capacitance is r Q 4 o = = π 1 1 R r oaxial able R 1

2 oaxial cable (two concentric conducting cylinders) of length L The inside conductor has a radius r with charge ρ and the inside surface of the outside conductor is R with charge - ρ λ R Q R a b = = ln = ln π o r π ol r The capacitance is Q π L o = = R ln r r R L Exercise There are electrical devices that are designed to store energy in this fashion These devices are referred to a "capacitors" To get an idea of the magnitude of the unit Farad, find how large a parallel plate capacitor must be in order to have a capacitance of one Farad Take the distance between the plates to be 01 mm apacitors in Electrical ircuits The circuit diagram of a capacitor You can "charge" a capacitor by connecting the capacitor to a battery (power supply) (Remember that in the electrostatic situation the wires (conductors) are uipotentials) ombinations of apacitors - this is necessary because capacitors with only certain values are available apacitors in parallel: The total capacitance of the circuit, that is uivalent to the capacitors is parallel (does the same job as the capacitors in parallel) + "top to top, bottom to bottom" "left to left, right to right" voltage "The voltage is the same across all capacitors in parallel" nd charge is _ conserved: = = = ; Q = Q + Q + Q 1 1 Using the definition of the capacitance: Q= = + + = + + ( ) = apacitors in Series: In this case the capacitors connected to each others "one after another" - similar to a train engine pulling its cars The total capabitance, can be obtained as follows "The charge on the capacitors that are in series is the same on each capacitor" voltage _

3 = + + ; Q = Q = Q = Q 1 1 Using the definition of the capacitance: Q1 Q Q = + + = Q = = + + Q 1 We can generalize our results for the total capacitances in the parallel and series circuits: = N N-apacitor connected in parallel = N-apacitor connected in series 1 N For two capacitors connected in series: + 1 = 1 Parallel and series combinations: Find the charge on each capacitor and the voltage across each capacitor Solution The uivalent capacitance of the circuit is = () Series( // 4) *6 = series(+ 4) = = µ F + 6 The total charge Q= =1*=4µ oulomb 1 v nd the charge on the µf capacitor is ual to the total charge: 4µ oulomb Potential(voltage) of this capacitor is =4/=8 olts From the conservation of energy voltages of the and 4µF capacitors are =1-8=4 volts Then the charge on the µf capacitor is 4*=8µ oulomb Then the charge on the µf capacitor is 4*=16µ oulomb Exercises Find the uivalent capacitance between points and 6 µf 4 µf µf µf 4 µf µf 8 µf µf

4 Find the uivalent capacitance between points and µf 6 µf µf 8 µf 4 µf µf and a 6 µf capacitor are connected in parallel and are charged by a 1 volt battery, as shown fter the capacitors are charged, the battery is then disconnected from the circuit The capacitors are then disconnected from each other and reconnected after the 6 µf capacitor is inverted Find the charge on each capacitor and the voltage across each 1 v µf 6 µf D D Energy stored in the capacitor When a capacitor is being "charged" by a battery (or power supply), work is done by the battery to move charge from one plate of the capacitor to the other plate s the capacitor is being charged, we can say that the capacitor is storing energy (What kind of energy?) Find the stored energy onsider a capacitor being charged by a battery fter a time t, the voltage across the capacitor is and an amount of charge q has accumulated (so far) on the plates of the capacitor To move an additional amount of charge dq from one plate to the other, the battery must do an amount of work dw, where dw = (dq) (Remember from before that W = a b U a U = b q( a b), or W = q is the work done moving a charge q through a voltage ) Formally: q Q q dw = du = dq = dq U dq = 0 = Q where U is the stored energy in the capacitor s a summary: U Q 1 1 = = = Q Question: Where is the energy stored? U Energy density is defined as the stored energy per unit volume: u = volume alculate the stored energy in a parallel plate capacitor of surface area and plate separation d Potential difference between the plates is 0 alculate energy density 0 Solution: apacitance of the parallel plate is: Q = = Then the energy is: d 1 0 U = 0 d In order to find energy density we divide U by volume =d; 4

5 U u = = 0 = 0E d d where E is the electric field between the plates and it can be defined as =Ed Dielectrics in capacitors careful glance at the uations of the capacitors shows that we can increase the capacitance of a capacitor by using some materials whose permittivity bigger than the permittivity of the air 0 These materials are known as the dielectric materials Dielectrics are insulators Electrons are not free to flow from one molecule to another The atoms in a dielectric can have dipole moments In a typical chunk of dielectric material these dipoles are randomly aligned Dipoles and therefore produce no net field as shown +Q Electric Field When a dielectric is placed between the plates of a capacitor with a surface charge density ρ the resulting electric field, E 0, tends to align the dipoles with s the field These results in a net charge density ρ s induced on the surfaces of the dielectric which in turns creates an induced electric field, E i, in the opposite direction to the applied field The total field inside the dielectric is reduced to, E = E0 Ei E0 The dielectric constant is defined as the ratio of the applied field to the total field, κ = E E0 1 (kappa) Substituting for E and solving for the induced field: κ = Ei = 1 E0 E0 Ei κ Note that κ=1 is a perfect insulator such as a vacuum and κ= is a perfect conductor How does the introduction of a dielectric affect the capacitance of a capacitor? We can find change in the potential: 1 1 = Ed = E0 d = 0 κ κ If the capacitance without dielectric 0 =Q/ 0, with dielectric it will be =Q/, eliminating Q, and 0 between uations we obtain = κ 0 -Q The capacitance larger by a factor κ Some values -- vacuum: κ = 1, glass: κ = 5 to 10, mica: κ = to 6 Find the capacitance of the capacitor shown in figure Solution Draw the circuit diagram of the capacitors These are parallel plate capacitors and their capavitance can be 0 calculated by using 0 = ; = κ0 Then: d / / κ 1 κ d/ κ d/ 5

6 0 / 0/ 0/ 1 = κ1 ; = κ ; = κ d d / d / Since and are series to the each other, the uivalent κκ 0 capacitance is: 1 = = + κ + κ d 1 is parallel to the 1 : κκ κκ κ 1 = 1+ 1 = + κ1 = + κ + κ d d d κ + κ onsider a parallel capacitor made of two large metal plates of L by L separated by distance d (<<) with a neutral dielectric slab Metal (thickness a, same area as the metal plates) The potential difference between the two plates is Find the amount of charge on the plates and energy stored in (a) and (b) L 1 Dielectric (a) Solution: x (a) We can think that two capacitor Metal Dielectric connected series with the capacitances 0 0L κ0l (b) 1 = = ; and = d a d a a The uivalent capacitance is 1 κ 0L = = 1+ a+ ( d a) κ The total charge on the plates Q= Energy stored in the capacitor is: W=Q/ (b) Equivalent circuit of the configuration is a capacitor connected in parallel to the two capacitoe in series with the capacitances: 0L( L x) 0Lx κ 1 ; ; 0Lx = = = d d a a The uivalent capacitance is = 1 + ; Q= and W=Q/ + Exercises parallel plate capacitor consists of plates of area 10 cm and a distance between the plates of 005 mm The space between the plates is filled with a dielectric of constant κ = 5 The capacitor is connected to a 6 volt battery a Find the capacitance of the capacitor with the dielectric b Find the charge on the plates of the capacitor c Find the induced charge on the surface of the dielectric d Find the energy stored in the capacitor e Find the energy density between the plates of the capacitor 6

7 Find the capacitance of the capacitor shown / / κ 1 κ d/ κ d/ 7