# Reading: Electrostatics 3. Key concepts: Capacitance, energy storage, dielectrics, energy in the E-field.

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1 Reading: Electrostatics 3. Key concepts: Capacitance, energy storage, dielectrics, energy in the E-field. 1.! Questions about charging and discharging capacitors. When an uncharged capacitor is connected across a battery, why do equal and opposite charges appear on the plates? (Why do they appear and why are they equal and opposite?) If the plates have different shapes or sizes, will the total charges still be equal and opposite? Explain. Where does the energy come from when a capacitor is charged by a battery? Where does it go if an isolated charged capacitor is discharged by connecting a small wire between the plates? [Make a guess.] 2.! Questions about altering the capacitance of a capacitor with parallel horizontal plates. Why does moving the plates closer together increase C? [Consider an isolated charged capacitor; what happens to ΔV?] If you suspend a copper sheet between the plates, parallel to them, what happens to C? Suppose the copper sheet is resting against one of the plates; is the change different? You have a copper sheet and a sheet of glass, the same size, that you can insert between the capacitor plates. Which one will change C more? Why? 3.! An isolated parallel plate capacitor has charge Q. If the plates are moved closer together, the stored energy decreases. Where does the lost energy go? If a dielectric slab is inserted part way into the gap between the plates and released, what happens? Explain. If, instead of being isolated, the capacitor remained connected to the charging battery, the energy increases when the plates are moved closer together. Explain this carefully in terms of how much work is done by what agent.

2 4.! Two capacitors C 1 and C 2 connected in series or parallel can be replaced by a single equivalent capacitor C, with charge Q and potential difference ΔV. If the two capacitors are in series, which of the two quantities Q and ΔV is the same for both and which is divided between them? Answer the same question if the two capacitors are in parallel. Derive the rules giving C in terms of C 1 and C 2 for the two cases. 5.! An external E-field produces aligned dipole moments in a dielectric material; this in turn results in the appearance of excess charge on the surfaces of the material. It is not free to move around, so it is called bound charge; let its area density be σ b. Consider a slab of dielectric filling the gap between the parallel plates of a charged and isolated capacitor. The charge on the plates would produce a field in the gap E 0 if the dielectric were not there. Show that the bound surface charge next to the positive plate of the capacitor is negative, and vice versa. [Draw a picture.] Express the net field in the slab, E av, in terms of E 0 and σ b. [Treat the bound charge distributions as infinite sheets of uniform charge.] Because the forces and torques that produce the aligned dipoles are proportional to E av, we can write σ b = ε 0 χe av, where the dimensionless number χ is the electric susceptibility of the material. Use this to show that E av = E 0 /(1 + χ). [The dielectric constant is defined as κ = 1 + χ.] 6.! Questions about energy in the E-field. The formula for energy density is u e = 1 2 κε 0 E2. The dielectric constant κ is greater than 1, but inserting a dielectric into a region where there is an E- field reduces the stored energy. Explain. Refer to Question 3 (a). Explain the reduction of stored energy in terms of energy in the E-field.

4 10.! A 1 µf parallel plate capacitor is connected to a 100 V battery. The separation between the plates is slowly reduced until the capacitance becomes 2 µf. What is the increase ΔU in the stored energy? Ans: J. How much additional charge ΔQ has been put on the plates? Ans: 10 4 C. How much additional energy was added to the system by the battery? [It moves charge ΔQ through potential difference 100 V.] Ans: 10 2 J. Account for the difference between your answers to (a) and (c) 11.! Consider the system shown of three identical parallel plate capacitors, each with capacitance C = 1 µf. 1 2 If 60 V is applied at the terminals and the capacitors become fully charged, how much energy is stored in capacitor #1? [Find the charge on it.] Ans: J. 3 The system is now isolated from the charging source, and a slab of dielectric with κ = 2 is slipped between the plates of #1, filling the gap completely. What is the new stored energy in #1? Ans: J. What is the new potential difference across the terminals? Ans: 40 V. 12.! The capacitor shown is half filled with a dielectric of constant κ. Its positive plate is at potential V 0 above that of its negative plate. The plates have total area A and are separated by distance d. What is the surface charge density σ on the bottom half of the positive plate? [Find E in the empty gap.] Ans: σ = ε 0 V 0 /d. + The potential difference between the plates is the same in both halves, because the top and bottom half of each plate is at the same potential. So the total E-field must be the same in both halves. Because of the dielectric the charge density σ on the top half of the plates must be larger than on the bottom half. What is σ? Ans: σ = κε 0 V 0 /d. Find the total charge on the positive plate and the capacitance of the system. Ans: Q tot = V 0 ε 0A(1 +κ ), C = ε 0A(1 +κ ). 2d 2d

5 13.! Two capacitors of capacitance C 1 and C 2 are arranged as shown. Initially charge Q is on capacitor #1 and #2 is uncharged. The switch shown is closed and after a time all the charges are again at rest. 1 2 What is the charge now on capacitor #1? Ans: Q 1 = Q C 1 /(C 1 + C 2 ). What is the potential difference across them? Ans: Q/(C 1 + C 2 ).! What is the change in stored energy after the switch is closed? Ans: ΔU = C 2 Q 2 C 1 2(C 1 + C 2 ).

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