Energy / Dielectrics in Capacitors

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1 Energy / Dielectrics in apacitors Polar dielectrics, like Water. In an Electric field Phys Lecture G. Rybka Q/ Q Q/ U Useful stuff Q Q

2 We ended last time looking at this clicker Q 0 3 Q 3 Q 3 Which of the following is NOT necessarily true: A) 0 B) total > ) 3 D) Q Q 3 E) + 3 is in parallel with battery (same ) total is in parallel with 3 ;; total + 3 Q: When might this be true? Q and Q 3 same since capacitors in series in parallel with combination of and 3 ;; same Δ

3 Reminder: Energy of a apacitor We obtain the energy stored in a charged capacitor by alculating the work provided (usually by a battery) to charge a capacitor to +/- Q: Incremental work dw needed to add dq to capacitor at voltage : - + Q/ q dw dq( ) dq The total work W to charge to Q is then given by: W Q qdq 0 Q We will use U as in Potential Energy Three equivalent forms: W U Q Q

4 licker Two identical parallel plate capacitors are connected to a battery. is then disconnected from the battery and the separation between the plates of both capacitors is doubled. d d d d What is the relation between U, the energy stored in, and U, the energy stored in? (a) U < U (b) U U (c) U > U After The charge on has not changed. The voltage on has not changed. U ½ Q /;; smaller;; U up U ½ ;; smaller;; U down Or, think that the work done to separate the plates with fixed charge in meant takes more Energy And, Energy in decreased. harge must leave in order to reduce the electric field so that the potential difference remains the same.

5 Questions Suppose the capacitor shown here is charged to Q and then the battery disconnected. d A Now suppose I pull the plates further apart so that the final separation is d. How do the quantities Q, U,, E, change? Q: U: : E: : remains the same.. no way for charge to leave. increases.. add energy to system by separating decreases.. Separation d increases;; A/d remains the same.. depends only on charge density increases.. equal to E x d, and d increases How much do these quantities change answers: d U d d U d d d

6 Dielectrics Empirical observation: Yes, it s an insulator Inserting a non-conducting material between the plates of a capacitor changes the ALUE of the capacitance. Definition: The dielectric constant of a material is the ratio of the capacitance when filled with the dielectric to that without it. i.e. 0 is always > (e.g., glass 5.6;; water 78)

7 Dielectric onstants Of arious Materials

8 Dielectrics By adding a dielectric you are just making a new capacitor with larger capacitance (factor of ) 0 Q Q A good thing because It is hard to make big capacitors with just air gaps Permits more energy to be stored than otherwise

9 What is going on? Parallel Plate Example harge a parallel plate capacitor filled with vacuum (air) to potential difference 0. Q 0 0 is deposited on each plate. E 0 σ/ε 0 Insert material with dielectric constant. harge Q remains constant Induced dipoles in material align Bulk middle is neutral Effective opposite E field added to original field gives SMALLER net E between plates: E E 0 / E d à smaller for same Q. Q/ à 0

10 What about GAUSS' LAW? How can field decrease if charge remains the same?? Answer: the dielectric becomes polarized in the presence of the field due to Q. E E 0» The molecules partially align with the field.» The field inside the dielectric (from the dipoles) opposes the original field and is responsible for the reduction in the effective field Rewrite Gauss' Law in presence of a dielectric:! E! ds q 0 ε0 ε 0 E ds q This form of Gauss' Law can be used in vacuum or dielectric alike where q represents the "free" charge.!!

11 licker Two parallel plate capacitors are identical except has half of the space between the plates filled with a material of dielectric constant. Both capacitors have charge Q ompare E, the electric field in the air of, and E, the electric field in the air of +Q E? -Q +Q E? -Q (a) E < E (b) E E (c) E > E The key here is to realize that the electric field in the air in must be equal to the electric field in the dielectric in!! The top plane is a conductor à equipotential surface. The bottom plane is a conductor à equipotential surface. is proportional to E d For this to happen, the charge density on each plane must be nonuniform to create equal electric fields!! Since >, for the same charge, <. onsequently, E < E.

12 heckpoint 8 Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After has been charged and disconnected, it is filled with a dielectric. The dielectric increases ;; > Q remains the same Q/ so decreases Alternately, recall E reduced à E E 0 /

13 heckpoint 0 Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After has been charged and disconnected, it is filled with a dielectric. A) B) ) Just learned > for same Q, so Recall Also, since Q unchanged, larger implies lower U

14 heckpoint The two capacitors are now connected to each other by wires as shown. How will the charge redistribute itself, if at all? A. The charges will flow so that the charge on will become equal to the charge on. B. The charges will flow so that the energy stored in will become equal to the energy stored in. The charges will flow so that the potential difference across will become the same as the potential difference across. D. No charges will flow. The charge on the capacitors will remain what it was before they were connected. must be the same! Q: U: Q Q Q Q U U U U

15 licker and Typical alculation x 0 0 x 0 /4 An air- gap capacitor, having capacitance 0 and width x 0 is connected to a battery of voltage. A dielectric ( ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor? First a licker: What changes when the dielectric added? A) Only B) only Q ) only D) and Q E) and Q Adding dielectric changes the physical capacitor does not change and changes changes Q changes

16 x 0 0 alculation Now, Strategic Analysis: alculate new capacitance Apply definition of capacitance to determine Q onsider to be two capacitances, and, in parallel x 0 /4 An air- gap capacitor, having capacitance 0 and width x 0 is connected to a battery of voltage. A dielectric ( ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor? Parallel plate capacitor: ε 0 A/d A 3 / 4 A 0 d d 0 3 / 4 (ε 0 A 0 /d 0 ) 3 / 4 0

17 alculation x 0 0 An air- gap capacitor, having capacitance 0 and width x 0 is connected to a battery of voltage. x 0 /4 A dielectric ( ) of width x 0 /4 is inserted into the gap as shown. 3 / 4 0 What is Q f, the final charge on the capacitor? Parallel plate capacitor filled with dielectric: ε 0 A/d A / 4 A 0 d d 0 ¼(ε 0 A 0 /d 0 ) / 4 0

18 alculation x 0 0 An air- gap capacitor, having capacitance 0 and width x 0 is connected to a battery of voltage. x 0 /4 A dielectric ( ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor? 3 / 4 0 / 4 0 parallel combination of and : + 0 ( 3 / 4 + / 4 )

19 alculation x 0 0 An air- gap capacitor, having capacitance 0 and width x 0 is connected to a battery of voltage. x 0 /4 A dielectric ( ) of width x 0 /4 is inserted into the gap as shown. 3 / 4 0 What is Q f, the final charge on the capacitor? / 4 0 What is Q? 0 ( 3 / 4 + / 4 ) Q Q f Q

20 Reminder: Where is the Energy Stored? Answer: in the Electric field itself onsider energy stored by a constant field in a parallel plate capacitor: U Q The Electric field is given by: E Q ( Aε / d ) The energy density u in the field is given by: σ Q Þ ε 0 ε 0 A U E ε0ad u U volume U Ad 0 ε0e Units: 3 J m

21 licker b a onsider two cylindrical capacitors, each of length L. has inner radius a and outer radius b. has inner radius a and outer radius b. If both capacitors are given the same amount of charge, what is the relation between U, the energy stored in, and U, the energy stored in? b a (a) U < U (b) U U (c) U > U U U Q Q U U ln ln ( b / a) ( b / a) Q πε0l b ln a

22 Energy expressions in apacitors (various forms) From: Q/ Q Q/

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