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Chapter 24 Capacitance and Dielectrics

Chapter 24 Capacitance and Dielectrics. Capacitance and dielectrics (sec. 24.1) Capacitors in series and parallel (sec. 24.2) Energy storage in capacitors and electric field energy (sec. 24.3) Dielectrics (sec. 24.4) Molecular model / polarization (sec. 24.5). C 2012 J. Becker.

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Chapter 24 Capacitance and Dielectrics

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  1. Chapter 24 Capacitance and Dielectrics • Capacitance and dielectrics (sec. 24.1) • Capacitors in series and parallel (sec. 24.2) • Energy storage in capacitors and electric field energy (sec. 24.3) • Dielectrics (sec. 24.4) • Molecular model / polarization (sec. 24.5) C 2012 J. Becker

  2. Learning Goals - we will learn: Ch 24 • The nature of capacitors, and how to calculate their ability to store charge.• How to analyze capacitors connected in a network.• How to calculate the amount of energy stored in a capacitor.• What dielectrics are, and how they make capacitors more effective.

  3. A "charged" capacitor can store charge. When a capacitor is being charged, negative charge is removed from one side of the capacitor and placed onto the other, leaving one side with a negative charge (-q) and the other side with a positive charge (+q). Any two conductors insulated from one another form a CAPACITOR.

  4. Q = C Vwhere C = eo A / dfor a parallel plate capacitor, where eo is the permittivity of the insulating material (dielectric) between plates.Recall that we used Gauss's Law to calculate the electric field (E) between the plates of a charged capacitor: E = s / eo where there is a vacuum between the plates.Vab = E d, so E = Vab /d A charged parallelplate capacitor. The unit of capacitance is called the Farad (F).

  5. 1 1 1 / Ceq = 1 / C1 + 1 / C2 Two capacitors in series and the equivalent capacitor.V = V1 + V2 and Q = C V

  6. Ceq = C1 + C2 Two capacitors in parallel and the equivalent capacitor.Q = Q1 + Q2 and Q = C V

  7. Capacitors can store charge and ENERGYDU = q DV and the potential V increases as the charge is placed on the plates (V = Q / C). Since the V changes as the Q is increased, we have to integrate over all the little charges “dq” being added to a plate:DU = q DVleads to U = òV dq = òq/C dq = 1/CòoQ q dq = Q2 / 2C. And using Q = C V, we getU = Q2 / 2C = C V2 / 2 = Q V / 2ENERGY STORED IN A CAPACITORThe potential energy stored in the system of positive charges that are separated from the negative charges is like a stretched spring that has potential energy associated with it.

  8. ELECTRIC FIELD ENERGY Here's another way to think of the energy stored in a charged capacitor: If we consider the space between the plates to contain the energy (equal to 1/2 C V2) we can calculate an energy DENSITY (Joules per volume). The volume between the plates is area x plate separation (A d). Then the energy density u isu = 1/2 C V2 / A d =eoE2 / 2 Substituting C =eoA / d and V = E du = 1/2 (eo A/d)(Ed)2 /Ad = eo E2/2 C 2012 J. F. Becker

  9. Energy density:u =eoE2 / 2 This is an important result because it tells us that empty space contains energy if there is an electric field (E) in the "empty" space. If we can get an electric field to travel (or propagate) we can send or transmit energy and information through empty space!!! C 201 2 J. F. Becker

  10. DIELECTRIC CONSTANT:K = C / Co = ratio of the capacitances V = Vo / K Effect of a dielectric between the plates of a parallel plate capacitor. Note – the charge is constant !

  11. A dielectric is added between the plates of a charged capacitor (battery not connected):Q = Qo, therefore Q = C V and Qo = Co VoCo Vo = C V, and if Vo decreases to V, Co must increase to C to keep equation balanced, and V = Vo Co/C Definition of DIELECTRIC CONSTANT:K = C / Co = ratio of the capacitances V = Vo / K C 2012 J. F. Becker

  12. Q24.18 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the energy stored in the capacitor? A. The stored energy increases. B. The stored energy remains the same. C. The stored energy decreases. D. not enough information given to decide

  13. The charges induced on the surface of the dielectric (insulator) reduce the electric field.

  14. “Polarization” of a dielectric in an electric field E gives rise to thin layers of bound charges on the dielectric’s surfaces, creating surface charge densities+si and –si.

  15. A neutral sphere B in the electric field of a charged sphere A is attracted to the charged sphere because of polarization.

  16. For each problem: Draw a clear and carefully labeled diagram. Write the necessary equations in terms of variables. Explicitly show all steps in calculations, including units.

  17. “Polarization” of a dielectric in an electric field E.

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