5/20/2011 Chapter 25 HITT An electron moves from point i to point f, in the direction of a uniform electric field. During this displacement: Capacitance-II In the last lecture: we calculated the capacitance C of a system of two isolated conductors. j i We also calculated the capacitance for some simple geometries. In this chapter we will cover the following topics: -Methods of connecting capacitors (in series , in parallel). -Equivalent capacitance. -Energy stored in a capacitor. -Behavior of an insulator (a.k.a. dielectric) when placed in the electric field created in the space between the plates of a capacitor. -Gauss’ law in the presence of dielectrics. A. the work done by the field is positive and the potential energy of the electron-field system increases B. the work done by the field is negative and the potential energy of the electron-field system increases C. the work done by the field is positive and the potential energy of the electron-field system decreases D. the work done by the field is negative and the potential energy of the electron-field system decreases E. the work done by the field is positive and the potential energy of the electron-field system does not change (25 - 1) Equivalent Capacitor Capacitors in parallel The fig.a we show three capacitors Consider the combination of capacitors shown in the figure to the left and to the right (upper part). We will substitute these combinations of capacitor with a connected in parallel. This means that Ceq = C1 + C2 + C3 single capacitor Ceq that is "electrically equivalent" to the capacitor group it substitutes. the plate of each capacitor is connected to the terminals of a battery of voltage V . We will substitute the parallel combination of fig.a with a single equivalent capacitor shown in fig.b which is also connected to an identical battery This means that if we apply the same voltage V across the capacitors in fig.a and fig.b (either right or left) by connecting The three capacitors have the same potential difference V across their plates. The charge on C1 is: q1 = C1V . The charge on C2 is: q2 = C2V . to a battery, the same charge q is provided by the battery. Alternatively, The charge on C3 is: q3 = C3V . The net charge q = q1 + q2 + q3 = ( C1 + C2 + C3 )V if we place the same charge q on plates of the capacitors in fig.a q ( C1 + C2 + C3 )V = = C1 + C2 + C3 V V For a parallel combination of n capacitors is given by the expression: and fig.b (either right or left), the voltage V across them is identical. The equivalent capacitance Ceq = This can be stated in the following manner: If we place the capacitor combination and the equivalent capacitor in separate black boxes, by doing electrical mesurements we cannot distinguish between the two. n (25 - 9) Ceq = C1 + C2 + ... + Cn = ∑ C j (25 - 10) j =1 Capacitors in series The fig.a we show three capacitors connected in series. This means that one capacitor is connected after the other. More complex capacitor systems In general a capacitor system The combination is connected to the terminals of a battery may consist of smaller capacitor of voltage V . We will substitute the series combination of fig.a with a single equivalent capacitor shown in fig.b groups that can be identified as connected "in parallel" or "in series" which is also connected to an identical battery. The three capacitors have the same charge q on their plates. The voltage across C1 is: V1 = q / C1. In the example of the figure C1 and C2 in fig.a are connected in parallel. The voltage across C2 is: V2 = q / C2 . They can be substituted by the equivalent capacitor C12 = C1 + C2 as The voltage across C3 is: V3 = q / C3 . shown in fig.b. Capacitors C12 and C3 in fig.b are connected in series. The net voltage across the combination V = V1 + V2 + V3 (25 - 11) 1 1 1 Thus we have: V = q + + C1 C2 C3 q q The equivalent capacitance Ceq = = → V 1 1 1 q + + 1 1 1 1 = + + C1 C2 C3 Ceq C1 C2 C3 They can be substituted by a single capacitor C123 as shown in fig.c C123 is given by the equation: 1 1 1 = + C123 C12 C3 (25 - 12) 1 5/20/2011 C = εo A/ d (a) C/2 (b)C (c) 2C area is doubled (d) 0 (e) Need more information 2 5/20/2011 A question What is the equivalent capacitance between the points A and B? A. 1 µF B. 2 µF C. 4 µF D. 10µF E. None of these A B What would a 10V battery do, i.e. how much charge will it provide, when it is connected across A and B? 40 µC -q' - q' + + dq' + + + V' Charge A q q' B Voltage O V' V the capacitor by assuming that we transfer a charge dq′ from the negative plate to the positive plate. We assume that the capacitor charge is q′ and the corresponding voltage V ′. The work dW required q′ for the charge transfer is given by: dW = V ′dq′ = dq′ C We continue this process till the capacitor charge is equal to q. The total work W = ∫ V ′dq′ = q W= 1 q′ 2 q2 = C 2 0 2C If we substitute q = CV we get: W = CV 2 2 q qV 2 Work W can also be calculated by determining the area A of triangle OAB which is Vq equal to ∫ V ′dq ′. Area = W = 2 (25 - 13) q -A - + + A + + + d U= εoE 2 Potential energy stored in a capacitor The work W spent to charge a capacitor is stored in the form of potential energy U = W that can be retrieved when is capacitor is discharged. Thus U = q 2 CV 2 qV = = 2C 2 2 Energy density 2 q CV = 2C 2 u= 1 q ′dq ′ C ∫0 or W = r E -q Energy stored an an electric field Consider a capacitor C which is has a charge q. We can calculate the work W required to charge 2 2 We can ask the question: where is the potential energy of a charged capacitor stored? The answer is counter intuitive. The energy is stored in the space between the capacitor plates where a uniform electric field E = V / d is generated by the capacitor charges. In other words the electric field can store energy in empty space! We define as energy densiry (symbol u ) the potential energy per unit volume. u = U V The volume V between the plates is: V = Ad where A is the plate area U CV 2 V2 ε V ε E2 = = = o = o Ad 2 Ad 2 Ad 2 d 2 This result, derived for the parallel plate capacitor holds in general 2 Thus the energy density u = (25 - 14) 3 5/20/2011 C = κ Cair C = κ Cair Capacitor with a dielectric q V -q q' V -q' V q -q V' q q In 1837 Michael Faraday investigated what happens to the capacitance C of a capacitor when the gap between the plates is completely filled with an insulator (a.k.a. dielectric) Faraday discovered that the new capacitance is given by : C = κ Cair Here Cair is the capacitance before the insertion of the dielectric between the plates. The factor κ is known as the dielectric constant of the material. Faraday's experiment can be carried out in two ways: 1. With the voltage V across the plates remaining constant In this case a battery remains connected to the plates . This is shown in fig.a 2. With the charge q of the plates remaining constant. In this case the plates are isolated from the battery This is shown in fig.b -q V -q q' V -q' V q -q V' q -q Fig.a : Capacitor voltage V remains constant This is bacause the battery remains connected to the plates After the dielectric is inserted between the capacitor plates the plate charge changes from q to q ′ = κ q q′ κq q The new capacitance C = = = κ = κCair V V V Fig.b : Capacitor charge q remains constant This is bacause the plates are isolated After the dielectric is inserted between the capacitor plates the plate voltage changes from V to V ′ = The new capacitance C = (25 - 16) Dielectrics : An atomic view In a region completely filled with an insulator of Dielectrics are classified into "polar" and "nonpolar" Polar dielectrics consist of molecules that have a non-zero dielectric constant κ , all electrostatic equations containing the constant ε o are to be modified by electric dipole moment even at zero electric field due to the asymmetric distribution of charge within the molecule (e.g. H 2O ). At zero electric field replacing ε o with κε o Example 1 : Electric field of a point charge inside a dielectric: E = conductor dielectric (see fig.a) the electric dipole moments are randomly r oriented. When an external electric field Eo is applied 1 q 4πκε o r 2 (see fig.b) the electric dipole moments tend to align r preferrentially along the direction of Eo because this configuration corresponds to a minimum of the potential Example 2 : The electric field outside an isolated conductor energy and thus is a position of stable equilibrium. Thermal random motion opposes the alignment and immersed in a dielectric becomes: E= κ q q q = = κ = κ Cair V′ V /κ V (25 - 15) q V σ κε o U = − pE cos θ (25 - 17) (25 - 18) S generates an internal electric field that apposes r r r Eo . Thus the net electric field E is weaker than Eo Gauss' law and dielectrics In chapter 23 we formulated Gauss' law assuming that the r r charges existed in vacuum. ε o ∫ E ⋅ dA = q or ε o Φ = q n̂ In this section we will write Gauss' law in a from which is suitable for cases in which dielectrics are present. Consider first the parallel plate capacitor shown in fig.a. q We will the Gaussian surface S. The flux Φ = Eo A = A nonpolal dielectric on the other hand consists of molecules that in the absence of εo an electric field have zero electric dipole moment (see fig.a) . If we place the dielectric r between the plates of a capacitor the external electric field Eo induces an electric dipole r r moment p that becomes aligned with Eo (see fig.b). The aligned molecules do not create any net charge inside the dielectric. A net charge appears at the left and right surfaces of the dielectric opposite to the capacitor plates. These charges come from negative and thus ordering is incomplete. Even so, the partial alignment produced by the exterlal electric field S n̂ → Eo = q εo A Now we fill the space between the plates with an insulator of dielectric constant κ (see fig.b) . positive ends of the electric dipoles. These induced surface charges have sign opposite We will apply Gauss' law for the same surface S. Inside S in addition to the plate charge q q − q′ we also have the induced charge q′ on the surface of the dielectric. Φ = EA = → to that of of the opposing plate charges. Thus the induced charges create an electric field r r E ′ which opposes the applied field Eo (see fig.c). As a result is that the r net electric field E between the capacitor plates is weaker. E E= o (25 - 19) E q − q′ q (eqs.1) From Faraday's experiments we have: E = o = (eqs.2) Aε o κ κ Aε o r r q If we compare eqs.1 with eqs.2 we have: q − q′ = → ε o ∫ κ E ⋅ dA = q (25 - 20) κ εo E= κ 4 5/20/2011 HITT Gauss' law in the presence of dielectrics r r ε o ∫ κ E ⋅ dA = q Even though the equation above was derived for the parallel plate capacitor is true in general. r Note 1 : The flux integral now involves κ E Note 2 : The charge q that is used is the plate charge, also known as "free charge" Using the equation above we can ignore the induced charge q′ Note 3 : The dielectric constant κ is kept inside the integral to desrcibe the most general case in which κ is not constant over the Gaussian surface A parallel-plate capacitor has a plate area of 0.2m2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 106 V/m between the plates, the magnitude of the charge on each plate should be: A. 8.9 x 10-7 C B. 1.8 x 10-6 C -6 C. 3.5 x 10 C D. 7.1 x 10-6 C -5 E. 1.4 x 10 C (25 - 21) Question • A parallel-plate capacitor has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about: • A. 0 B. 5N C. 9N D. 1 x104 N E. 9 x 105 N Question • A. B. C. D. E. A parallel-plate capacitor has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about: 0 5N 0. 9N 1 x104 N 9 x 105 N F = qE = = q2 2ε o A (5 × 10 ) −6 2 2 × 8.85 × 10 −12 × 0.3 = 4.71N The electric field = σ/2εo why? 5 5/20/2011 Summary • Capacitor networks; Series, parallel and other more complex configurations. Equivalent capacitance. • Energy stored on a capacitor • Dielectric between the capacitor plates. – Dielectric constant and dielectric strength – Polar and non-polar molecules 6