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Lecture 7 Happy Groundhog Day!! Punxsutawney Phil saw his shadow this morning…6 more weeks of winter! Energy Storage in Capacitors The electric potential energy, U, stored in a capacitor is equal to the amount of work required to charge it. (work is required because of the force that charges exert on each other.) Let’s define U = 0 to be the electric potential energy of an uncharged capacitor (Q = 0, V = 0). When the capacitor is fully charged, it will have a charge Q and potential difference V = Q/C. At some intermediate point during charging, when the capacitor has a charge q, the potential difference will be V = q/C. To increase the charge on the capacitor by a small amount dq will then require an amount of work, dW q dW = ΔVdq = dq C To find the total work required to charge the capacitor, we need to integrate from q = 0 to q = Q: W =∫ Q 0 q Q2 dq = C 2C Energy storage in capacitors (cont.) Thus, the electric potential energy stored in a charged capacitor is: 1 Q2 U= 2 C • Larger capacitance => less energy is required to store a fixed charge Q 1 U = CV 2 2 Since Q = CV , we can also express this as: • Larger capacitance => greater ability to store energy at given potential difference. • Analogous to elastic potential energy: U = 1 kx (P7E) 2 2 1 U = QV Using V = Q / C , we can re-write the same expression as: 2 • This can be thought of as the total charge Q multiplied by the average potential difference (V/2) during the process of charging. Electric field energy The process of charging a capacitor creates an electric field between the conductors •Wherever there is a potential difference, there is an electric field We can think of the energy stored in a capacitor as being stored in the electric field between the plates. For a parallel plate capacitor, the energy density u of the electric field is defined as the stored energy divided by the volume between the plates. If the plate area is A and plate separation is d: A For a parallel plate capacitor, C = ε 0 d 1 CV 2 U U u= = = 2 Volume Ad Ad and V = Ed : 1 2 1 CV 2 ε 0 A d (Ed ) 1 u= 2 =2 = ε 0 E 2 [J/m 3 ] Ad Ad 2 ( ) Electric field energy The energy density: 1 u = ε 0 E 2 [J/m 3 ] 2 for the special case of a parallel plate capacitor is actually much more general and describes the energy density of any electric field in vacuum. Thus we can think of the electric potential energy as being associated with the arrangement of the charges, or as being stored in the electric field those charges create. Although we think of a vacuum as being “empty”, if it contains an electric field, it isn’t really empty after all, since it contains energy. Later in the course, we will use this relationship in connection with electromagnetic waves, which can transport energy through a vacuum. Quiz: Energy stored in capacitor You reposition the two plates of a capacitor so that the capacitance doubles. There is vacuum between the plates. If the charges +Q and –Q on the two plates are kept constant in this process, the energy stored in the capacitor… A.becomes 4 times greater. B.becomes twice as great. C.remains the same. D.becomes 1/2 as great. E.becomes 1/4 as great. 1 U = CV 2 2 1 Q2 U= 2 C 1 U = QV 2 Quiz: Energy stored in capacitor You reposition the two plates of a capacitor so that the capacitance doubles. There is vacuum between the plates. If the charges +Q and –Q on the two plates are kept constant in this process, the energy stored in the capacitor… A.becomes 4 times greater. B.becomes twice as great. C.remains the same. D.becomes 1/2 as great. E.becomes 1/4 as great. 1 Q2 U= 2 C 1 Q2 U= 2 C 1 Q2 U' = 2 2C Dielectrics What happens if we fill the region between the plates with a dielectric? A: The potential difference between the plates decreases. ➡ Suppose the potential decreases by a factor K > 1, so V=V0/K. With a vacuum between the plates, if the charge stored is Q and the potential difference is V0, the capacitance is C0 = Q/V0 With a dielectric between the plates, if the charge stored is Q and the potential difference is V < V0, the capacitance is: C= Dielectric const. K= C C0 Q Q = K = KC0 V V0 For Q=const. V= V0 K C = KC0 Dielectric constants The factor K depends on the material we insert, and is called the “dielectric constant” Vacuum: C =1. Air: C ~ 1. For other materials, K > 1: Induced Charge and Dielectric Polarization If a dielectric reduces the potential difference between two plates with fixed charge by a factor K, it must also reduce the electric field by the same factor: E0 E= K for constant Q Why? Electric field of the charges on the plates polarizes the dielectric material, creating an induced charge on the surfaces. •The charges inside the material of the dielectric rearrange themselves due to the presence of the field E0 •The induced charges on the surface of the dielectric create an induced field opposite to E0 •The induced field partially cancels E0, so the total field inside the dielectric is less than E0 Induced polarization is the effect that dominates in most materials used as dielectrics in capacitors Dielectric Permittivity The product ε = Kε 0 is called the permittivity of the dielectric. Using the permittivity, the electric field in the dielectric can be expressed in the same form as for vacuum: σ E= ε Similarly, for a parallel plate capacitor with a dielectric: C = KC0 = Kε 0 A A =ε d d And the energy density of the electric field in a dielectric is: 1 u = εE 2 2 Induced Charge For most materials, and moderate field strengths, the induced surface charge density of the dielectric is proportional to the original field E0 •We will assume that is true If the charge density on the plates is σ and the induced charge density on the surface of the dielectric is σi, the net surface charge density, σnet, on each side of the capacitor is σ – σi Since the electric field is given by: • E0 = σ ε 0 • E= σ − σ i E0 = ε0 K Solving for σi: E= σ net we have ε0 1⎞ ⎛ σ i = σ ⎜ 1 − ⎟ induced surface charge density ⎝ K⎠ If K is very large, σi≅σ, and thus the net electric field and voltage are much reduced. Properties of Capacitor Vacuum Electric field σ E= ε σ E0 = ε0 Capacitance A C0 = ε 0 d Electric energy density 1 u0 = ε 0 E02 2 With dielectric Q C= Vab A C=ε d 1 u = εE 2 2 Dielectric constant K= ε0: permittivity of free space ε = Kε 0 In dielectric replace ε0 with ε Permittivity C V0 = C0 V Brief Announcement — Ch 24 homework will be assigned asap, due next week. Chapter 25 Current & Resistance Electric Current ● ● Electric current is the rate of flow of charge through some region of space (or a wire). The SI unit of current is the ampere (A) ● ● 1A=1C/s The symbol for electric current is I dQ I= dt Direction of Current ● ● ● The charges passing through the area could be positive or negative or both It is conventional to assign to the current the same direction as the flow of positive charges Convention: direction of current flow is opposite the direction of the flow of electrons #thanksbenfranklin Charge Motion in a Conductor ● Electric field forces cause the electrons to move in the wire and create a current • Most electron motion in a conductor is random, caused by collisions. • The net “drift velocity” is very slow. Current and Drift Speed ̣ ̣ ̣ ̣ Wire of cross-section area A n = charge carriers per Volume nAΔx = number of charges that move through A in time Δt Total charge is ΔQ = (nAΔx)e Drift Velocity dQ I= = n q Avd dt Electron drift is SLOW Current: I ' 1 A = 1 C/s Density of electrons in wire: Electron charge: Area of wire: A ⇠ 10 Definition of current: v= q ⇠ 10 6 n ⇠ 1029 /m3 19 C m2 Drift Velocity dQ I= = nqAv dt I 1C/s = 29 nqA 10 · 10 19 · 10 6 C/m = 10 4 m/s wikipedia slower than a snail faster than continental drift Why is the current flow nearly instantaneous? If drift velocities are ~10-4 m/s, why do the lights turn on instantly? Answer: a conductor has mobile charges all along its length. Turning on a switch establishes an electric field through the conductor almost instantaneously, which means all mobile charges start drifting immediately. •A charge doesn’t need to travel from the switch to the light-bulb for the light-bulb to turn on – there are already charges inside the light-bulb, and the light-bulb turns on as soon as they start drifting, which happens almost instantly. Current and Work We just showed that the kinetic energy of the charges does not increase appreciably. So Where does the energy go? •Due to the frequent collisions of the moving charges, the energy is transferred to the (stationary) ions of the material, increasing their vibrational energy, and thus their temperature. Bulk of the work done by the electric field goes into heating the conductor. (this is ‘wasted’ energy, lots of room to increase efficiency) Quiz: current These four wires are made of the same metal. Rank in order, from largest to smallest, the electron currents ia to id: A.id>ia>ib>ic B.ib=id>ia=ic C.ic>ib>ia>id D.ic>ia=ib>id E.ib=ic>ia=id dQ I= = n q Avd dt Quiz: current These four wires are made of the same metal. Rank in order, from largest to smallest, the electron currents ia to id: A.id>ia>ib>ic dQ I= = n q Avd dt B.ib=id>ia=ic C.ic>ib>ia>id D.ic>ia=ib>id E.ib=ic>ia=id (πr )v 2 (πr )2v 2 (π 4r )v 2 ⎛ 1 2⎞ ⎜⎝ π r ⎟⎠ 2v 4 Quiz: current and drift speed Two copper wires of different diameter are joined end to end, and a current flows in the wire combination. When electrons move from the larger-diameter wire into the smaller-diameter wire, A.their drift speed increases. B.their drift speed decreases. C.their drift speed stays the same. D.not enough information given to decide Quiz: current and drift speed Two copper wires of different diameter are joined end to end, and a current flows in the wire combination. When electrons move from the larger-diameter wire into the smaller-diameter wire, A.their drift speed increases. B.their drift speed decreases. C.their drift speed stays the same. D.not enough information given to decide Analogy: water is a pipe (when radius of pipe decreases, the speed increases). The current I is constant for the whole wire. If A decreases, v has to decrease to compensate. dQ I= = nqAv dt Current density Current: dQ I= = n q Avd dt Take out the area factor => current density: J = I = n q vd ⎡ A 2 ⎤ ⎣ m⎦ A In some cases, it is useful to treat the current density as a vector: • q > 0, vd same direction as E • q < 0, vd opposite direction as E Total current through a surface: J describes how charges flow at a certain point • Direction of J changes around the circuit • Magnitude of J = I/A can change along the circuit (e.g. when A changes) Conductivity ● Currents flow in response to electric fields. ● For some materials (e.g. metals), the current density is [nearly] directly proportional to the electric field ● The constant of proportionality, σ, is called the conductivity of the conductor J=σE “Ohm’s Law” current density: Ohm’s Law ● ● ● ● I = n q vd A Ohm’s law : for many materials, the current density is proportional to the electric field: J = σ E Most metals obey Ohm’s law - Materials that obey Ohm’s law are called ohmic Not all materials follow Ohm’s law ● ● J=σE J= Materials that do not obey Ohm’s law are said to be nonohmic Ohm’s law is not a fundamental law of nature Ohm’s law is an empirical relationship valid only for certain materials ● Specifically when conduction properties are determined by collisions of electrons with atoms ● ● ● ● Ohm (1789 -1854) German physicist Formulated idea of resistance Discovered Ohm’s Law