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Parallel Plate Capacitor

Parallel Plate Capacitor
Capacitors II Physics 2415 Lecture 9 Michael Fowler, UVa Today’s Topics • First, some review… then • Storing energy in a capacitor • How energy is stored in the electric field • Dielectrics: why they strengthen capacitors Parallel Plate Capacitor • a • E = / = Q/A and V = Ed, 0 0 area A d apart Qd Q • so V AC 0 A Q where C 0 + + + + + + + d d apart Know this formula Q Charge will settle on inside surfaces Capacitors in Parallel • Let’s look first at hooking up two identical parallel plate capacitors in parallel: that a means the wires from the two top plates are joined, similarly at the bottom, so effectively they become one capacitor. • What is its capacitance From the picture, the combined capacitor has twice the area of plates, the same distance apart. • We see that C = C + C 1 2 C C 1 2 Capacitors in Parallel • If two capacitors C , C are wired together as 1 2 shown they have the same voltage V between plates. • Hence they hold charges Q = C V, Q = C V, 1 1 2 2 for total charge Q = Q + Q = (C + C )V = CV. 1 2 1 2 • So capacitors in parallel just add: C = C + C + C + … 1 2 3Q Q Q Q A Capacitors in Series C C 2 1 • Regarding the above as a single capacitor, the important thing to realize is that in adding charge via the outside end wires, no charge is added to the central section labeled A: it’s isolated by the gaps between the plates. • Charge Q on the outside plate of C will 1 attract –Q to the other plate, this has to come from C , as shown. 2 • Series capacitors all hold the same charge. Q Q Q Q Capacitors in Series A C C 2 1 • Series capacitors all hold the same charge. • The voltage drop V across C is V = Q/C . 1 1 1 1 • The voltage drop across C is V =Q/C . 2 2 2 • Denoting the total capacitance of the two taken together as C, then the total voltage drop is V = Q/C. • But V = V + V , so Q/C = Q/C + Q/C , 1 2 1 2 1 1 1 Important  C C C 12It Takes Work to Charge a Capacitor • Suppose a sphere capacitor C already contains charge q. Then it’s at a potential V(q) = q/C. • To bring a further little charge dq from far away, against the repulsive force of the charge already there, takes work V(q)dq. • So, to deliver a total charge Q to the capacitor, one bit dq at a time, takes total work: QQ 2 qQ W V q dq dq   CC 2 00Work Done in Charging a Parallel Plate Capacitor • q • The math is identical to charging a + sphere by bringing up little charges + dq from far away (see the last slide) but for + the parallel plate capacitor we only + have to bring charge across from one + q q plate to another: the work is still + V(q)dq for each dq. + + • A capacitor is actually charged, of + course, by using a battery to pump + charge from one plate to the other via + an outside wire, but the route taken + doesn’t affect the gain in potential V(q) = q/C energy of the charge transferred. Energy Stored in a Capacitor • The work needed to place charge in a capacitor is stored as electrostatic potential energy in the capacitor: 2 Q 2 11 U CV QV 22 2C • We proved the first, the others come from V = Q/C. Clicker Question • How much work does it take to put charge Q on to a 2mF capacitor compared with putting the same charge on to a 1mF capacitor A. Twice as much B. The same C. Half as much Clicker Answer • How much work does it take to put charge Q on to a 2mF capacitor compared with putting the same charge on to a 1mF capacitor • Half as much: the 2mF capacitor is only at half the voltage of the 1mF capacitor if they have the same charge q, so bringing up extra charge dq takes only half the work. Pulling the Plates Apart • Suppose we have charge Q on a • a parallel plate capacitor having area A and plate separation d. • We now pull the plates to a greater distance apart, say 2d. • Assume first that the capacitor is disconnected from the battery, so no charge can flow. • Since the plates are oppositely charged, it takes work to pull them apart. Reminder from lecture 3 Field for Two Oppositely Charged Planes • a = + Superpose the field lines from the negatively charged plate on the parallel positively charged one, and you’ll see the total field is double in the space between the plates, but exactly zero outside the plates. Working to Pull the Plates Apart • From the last slide, the electric field • a E = / between the plates is /2 0 0 from the top plate and /2 from 0 the bottom plate. • Therefore, in finding the work done against the electric field in moving the top plate, charge Q, we can only count the field from the bottom plate—a charge can’t do work moving in its own field Working to Pull the Plates Apart • Moving the top plate, charge Q = A, a x •• a a distance x outwards in the electric field /2 from the bottom plate takes work 0 2  Qx 2 1 W Fx Q x x Q 2 22AA 0 0 0 1 d 1 dx  • Now initially , finally CA CA 0 0 so work done = capacitor energy change: 22 QQ W 22 CC final initialWorking to Pull the Plates Apart • Moving the top plate, charge Q, a x • a distance x outwards created an extra volume V = Ax between the plates, filled with the constant electric field E, and took work: 2 Q Ax 2 1 W x EV 0 2 22  00 • The electric field itself is the store of energy: and this is true in general, for varying as well as constant fields, the 2 1  E . energy density in an electric field is 0 2Clicker Question • Suppose the parallel plates are pulled • a apart from separation d to 2d, the plates having a constant potential V difference V from a battery. • What happens to the electric field strength between the plates A. It’s doubled B. It’s halved V C. It’s constant Clicker Answer • Suppose the parallel plates are pulled • a apart from separation d to 2d, the plates having a constant potential V difference V from a battery. • What happens to the electric field strength between the plates A. It’s doubled B. It’s halved V same voltage, double distance: V/m. Clicker Question • Suppose the parallel plates are pulled • a apart from separation d to 2d, the plates having a constant potential difference V from a battery. V • What happens to the total energy stored in the capacitor A. It’s doubled B. It’s halved C. It’s constant V D. It increases by a factor of 4 E. It decreases by a factor of 4 Clicker Answer • Suppose the parallel plates are pulled • a apart from separation d to 2d, the plates having a constant potential difference V from a battery. V • What happens to the total energy stored in the capacitor A. It’s doubled 2 1 B. It’s halved U CV 2 C. It’s constant V D. It increases by a factor of 4 E. It decreases by a factor of 4 Puzzle… • a • Pulling the plates apart takes external work, since the plates are V oppositely charged and attract each other. • Yet after pulling them to double the initial separation, there is less energy stored in the capacitor than before V • What about conservation of energy Puzzle Answer • a • What about conservation of energy • The capacitance goes down, the V voltage is constant, so charge flows from the capacitance into the battery. (Q = CV) • And, it flows the wrong way—against the battery’s potential, so this takes work. The battery is being charged, V it’s storing energy. Field Energy for a Charged Sphere • For a charged spherical conductor of radius R: ˆ 11 Qr Q E r and V r .  2 44rr 00 • The energy stored in the electric field is 2  2  QQ 22 11 U E dv 4r dr  00 22 2  48rR 00 R 1 and this is just , so the capacitor’s energy is QV 2 in the electric field. How Big is an Electron • We’ve just seen that a charge Q on a sphere of radius R has electric field energy 2 Q U 8  R 0 • This means that since the electron has a charge, if the inverse square law holds up at smaller and smaller distances, it can’t be infinitely small • A lower limit on its size is given by assuming its mass comes entirely from this electrostatic energy, using 2 U = E = mc . 15 • This gives R about 10 m: called the classical radius of the electron. In fact, at this R, Coulomb’s law breaks down—and we need quantum mechanics. Dielectrics • If a nonconducting material is • a placed in an electric field, the + + + electrons will still move a little, + + + remaining within their home + + + molecules or atoms, which will + + + therefore become polar. + + + + + + • The overall effect of this polarization is to generate a layer of positive charge on the right and negative charge on the left. Dielectrics + + + • If a nonconducting material is • a + + + placed in an electric field, the + + + + + + electrons will still move a little, + + + remaining within their home + + + molecules or atoms, which will therefore become polar. + • The overall effect of this + polarization is to generate a + layer of positive charge on the + + right and negative charge on + the left. Dielectrics + + + • These “layers of surface excess • a + + + charge” created by the + + + + + + polarization generate an electric + + + field opposing the external + + + field. • However, unlike a conductor, + this field cannot be strong + enough to give zero field inside, + because then the polarization + + would all go away. +