Energy storage in CAPACITORs

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Martina Strickland
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1 Energy storage in CAPACITORs Charge capacitor by transferring bits of charge dq at a timefrom bottom to top plate. Can use a battery to do this. Battery does work which increase potential energy of capacitor. q is magnitude of charge on plates V= q/c V across plates du = V dq increase in potential energy +q -q V dq U U du Q q dq Q CV 1 C C C 0 0 CV Y&F, eqn. 4.9 two ways to write

2 Where is the Energy Stored? Claim: energy is stored in the electric field itself. Think of the energy needed to charge the capacitor as being the energy needed to create the field. To calculate the energy density in the field, first consider the constant field generated by a parallel plate capacitor, where Q Q The electric field is given by: 1 Q 1 Q C ( A / d) Q E A U 0 E Ad The energy density u in the field is given by: U U U 1 u 0E volume Ad 0 This is the energy density, u, of the electric field. Units: J 3 m

3 Example Energy Density Claim: the expression for the energy density of the electrostatic field 1 u 0 E is general and is not restricted to the special case of the constant field in a parallel plate capacitor. Consider E- field between surfaces of cylindrical capacitor: Calculate the energy in the field of the capacitor by integrating the above energy density over the volume of the space between cylinders. 1 1 U 0 E dv 0 E r dr dl etc. Compare this value with what you expect from the general expression: 1 W CV

4 Capacitor Summary A Capacitor is an object with two spatially separated conducting surfaces. The definition of the capacitance of such an object is: C Q V The capacitance depends on the geometry : d A r +Q a b -Q L -Q +Q a b Parallel Plates A d C 0 C Cylindrical 0 L b ln a C Spherical 4 b 0 ab a

5 Example 1 Consider two cylindrical capacitors, each of length L. C 1 has inner radius 1 cm and outer radius 1.1cm. C has inner radius 1 cm and outer radius 1.cm. If both capacitors are given the same amount of charge, what is the relation between U 1, the energy 1. stored in C 1, and U, the energy stored in C? C 1 C (a) U < U 1 (b) U = U 1 (c) U > U 1

6 Example 1 Consider two cylindrical capacitors, each of length L. C 1 has inner radius 1 cm and outer radius 1.1cm. C has inner radius 1 cm and outer radius 1.cm. If both capacitors are given the same amount of charge, what is the relation between U 1, the energy 1. stored in C 1, and U, the energy stored in C? C 1 C C (a) U < U 1 (b) U = U 1 (c) U > U 1 The magnitude of the electric field from r = 1 to 1.1 cm is the same for C 1 and C. But C also has electric energy density in the volume 1.1 to 1. cm. In formulas: ol 1 C1 ~ r 1.1 C 1 ~ U 1. Q / C C1 1. ln( ) outer ln ln ln U1 Q / C1 C r inner

7 DIELECTRICS Consider parallel plate capacitor with vacuum separating plates (left) Suppose we place a material called a dielectric in between the plates (right) The charge on the plates remain the same, but a dielectric has a property of having induced charges on its surface that REDUCE the electric field in between and the voltage difference. Since C = Q/V, the resulting capacitance will INCREASE. Y&F Figure 4.13

8 DIELECTRICS Suppose the charges on the plate and the dielectric are, s and s i. The electric Fields before and after are E E i K 0 0 ; ; 0 0 E E We define the ratio of the original field over the new field as the dielectric constant, K. Hence, the voltage difference changes by 1/K and the capacitance, C o =Q/V, changes by C=KQ/V=K C o i For same Q: C = KC o E = E o /K V = Vo/K But C = KC o General

9 DIELECTRICS Materials Glass, mica, plastics are very good dielectrics

10 DIELECTRICS and permittivity We introduce a convenient redefinition of ε 0, called permittivity, as 0 Consider a parallel plate capacitor with no dielectric C0 0 A capacitor with a dielectric becomes simply, C A d KC0 K0 A d A d The change in capacitance can be accounted for by changing permittivity.

11 EXAMPLE of parallel plate capacitor problem A parallel plate capacitor is made by placing polyethylene (K =.3) between two sheets of aluminum foil. The area of each sheet is 400 cm, and the thickness of the polyethylene is 0.3 mm. Find the capacitance. C =K ε o A/d = (.3) (8.85 x 10-1 C /Nm ) (400 cm )(1m /10 4 cm ) 0.3 x 10-3 m =.71 nf

12 Example : Two identical parallel plate capacitors are connected to a battery. Remaining connected, C is filled with a dielectric. Compare the voltages of the two capacitors. a) V 1 > V b) V 1 = V c) V 1 < V

13 Example : Two identical parallel plate capacitors are connected to a battery. Remaining connected, C is filled with a dielectric. Compare the voltages of the two capacitors. a) V 1 > V b) V 1 = V c) V 1 < V

14 Example 3: Two identical parallel plate capacitors are connected to a battery. Remaining connected, C is filled with a dielectric. Compare the charges on the plates of the capacitors. a) Q 1 > Q b) Q 1 = Q c) Q 1 < Q

15 Example 3: Two identical parallel plate capacitors are connected to a battery. Remaining connected, C is filled with a dielectric. Compare the charges on the plates of the capacitors. a) Q 1 > Q b) Q 1 = Q c) Q 1 < Q Note: Unlike constant Q case, here V and E remain the same but C = K C o still.

16 EXAMPLE Two parallel plate capacitors, C 1 = C = μf, are connected across a 1 V battery in parallel. a.) What energy is stored? 1 U1 U CV 144J UT 88J b.) A dielectric (K =.5) is inserted between the plates of C. Energy? C U ' ' KC 1 ' C V.5 F 360J 5F U T 504J Note: a dielectric increases amount of energy stored in C.

17 Y&F Problems 4.7 and 4.71 A parallel plate capacitor has two dielectrics, side by side, show the capacitance is, C 0 A d K 1 K A parallel plate capacitor has two dielectrics, stacked, show the capacitance is, C 0 A d K1K K K 1

18 More weekend Fun HW #4 get cracking (Hints on Monday) Office Hours immediately after this class (9:30 10:00) in WAT14 [1-1:30pm today] nd Quiz Now

HW7 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case.

HW7 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 24.P.021 (a) Find the energy stored in a 20.00 nf capacitor