q = CV = 25nF $ 120V = 3000nC

Transcription

1 hapter 5 5. The capacitor in Fig. 5-6 has a capacitance of 5nF an is initially uncharge. The battery provies a potential ifference of 0 V. After switch S is close, how much charge will pass through it. This question is really asking what the charge on the is... q V 5nF $ 0V 3000n 5.4 You have two flat metal plates, each of area.00m with which to construct a a parallel plate capacitor. (a) If the capacitance of the evice is to be.00 F, what must be the separation between the plates. (b) oul this capacitor actually be constructe......no way you coul o this A A m.00f m 5.7 What is the capacitance of a rop that results when two mercury spheres, each of raius R.00mm, merge? V new $ 3 4 rr3 6.7 # 0-8 m 3 r new.5 # 0 - m 4rf 0r.8 # 0-3 F We first calculate the volume of the resulting rop. This allows us to fin the raius of the resulting rop 5.8 In Fig. 5-7 fin the equivalent capacitance of the combination. Assume that 0µ F, 5µ F an 3 4µ F. apacitors an are in series. We calculate the equivalent capacitance for this combination an then note that it is in parallel with 3.

2 + series 0µF + 5µF 3 0µF series 0 3 µf parallel 0 3 µf + 4µF 3 µf 5.9 In Fig. 5-9 fin the equivalent capacitance of the combination. Assume that 0µ F, 5µ F an 3 4µ F. apacitors an are in parallel. We calculate the equivalent capacitance for this combination an then note that it is in series with 3. parallel + 5µF series 5µF + 4µF 4 60µF µF 9 60µF series 60 9 µf 5. Each of the uncharge capacitors in Figure 5-30 has a capacitance of 5nF. A potential ifference of V 400 V is establishe when the switch is close. How many coulombs of charge pass through meter A. The charge that passes through the meter is the charge that woul en up on the equivalent capacitor. We first fin the equivalent capacitance... eq 5nF + 5nF + 5nF 75nF We can now compute the charge on the equivalent capacitor q eq eq V 75 # 0-6 F $ 400V In Fig. 5-30, the battery has a potential ifference of V0 V an the five capacitors each have a capacitance of 0 µ F. What is the charge on (a) capacitor an (b) capacitor. We can calculate the charge on capacitor one very easily, since we know the voltage across it. q V 0.0µ F 0.0V 00µ

3 To tackle the charge on, we nee to o some substitutions. The picture below shows the successive substitutions. eq eq 3 e q eq + eq 5µ F eq 3 eq + 5µ F +0µ F 5µ F + eq eq 5µ F + 0µ F 6µ F eq Now that we know the progression of substitutions, we can start with the final equivalent capacitor an work backwar to fin the charge on. We know that the voltage across the equivalent capacitor is 0V. q eq eq V 6µF 0V 60µ The charge on eq3 is the same as the charge on eq since eq arose from a series replacement. The charge on series capacitors an their replacement is always the same. This allows us to fin the voltage across eq3. V eq 3 q eq 3 eq 3 60µ 5µ F 4V This is the voltage across eq since parallel capacitors an their equivalent replacement always have the same voltage. This allows us to compute the charge on eq. q eq eq V eq 5µ F 4V 0µ

4 This is the charge on. We know this because the charge on series capacitors an their replacement is the same.. q 0µ 5.7 Di in problem session (I think). 5.3 In Fig. 5-39, the battery has potential ifference V9.0, 3.0µF, 4 4.0µF, an all the capacitors are initially uncharge. When switch S is close, a total charge of µ passes through point a an a total charge 8µ passes through point b. What are (a) an (b) 3? We know that the charge on 4 must be 8µ since all of that charge passe through point b. We can calculate the voltage on 4 V 4 q 4 4 8µ 4µF V If µ passe through a an 8µ passe through (b), we know that 4µ must be on 3. Using this charge an the voltage we compute, we can fin 3. 3 q 3 4µ V 3 V µ F

5 V a b 3 4 Since µ passe through a, we know that the charge on both an must be µ. We can compute the voltage across. V q µ 3µF 4V We know that the voltage across all of the capacitors is 9V, an the voltage across an 4 are 4V an V respectively. This means that the remaining 3V rop occurs across. Using the charge an voltage, we can now compute. q µ V 3V 4 µ F 5.6 Figure 5-4 isplays a.0 V battery an 3 uncharge capacitors of capacitances 4.00nF, 6.00nF, an nF. The switch is thrown to the left sie until capacitor is fully charge. Then the switch is thrown to the right. What is the final charge on (a) capacitor, (b) capacitor, an (c) capacitor 3? Initially we charge capacitor by itself. It is straightforwar how to compute its charge. q init V 4.00nF $ V 48 n Now that we know the charge on capacitor, we can flip the switch. When we o this, charge will flow from to an 3. At the en, the voltage across capacitor will equal the sum of the voltages across capacitors an 3

6 V final V final + V 3final q final q final q 3final + 3 Series apacitors: q final q 3final q q final q q + 3 q final q a + 3 k q final q a + 3 k harge onservation: q init q final + q q init q final + q q a + 3 k + q q init q; a + 3 k + E q 3 + 9a 3 k q init q q q init 3 6 nf $ 3 nf q ; 4 nf $ 6 nf + 6 nf $ 3 nf + 4 nf $ 3 nf E48 n q 6 n q final q init - q 3n 5.9 A.0 µf an a 4.0 µf are connecte in parallel across a 300V potential ifference. alculate the total energy store in the capacitors. U V + V 0 6 (300V ) (300V ) 0.7J 5.3 What capacitance is require to store an energy for 0 kwh at a potential ifference of 000 V. We convert to Joules first, an then fin. U 0 kwh 0,000W $ 3600s 3.6 # 0 7 J U V U V $ 3.6 # 0 7 J 7 F ^000Vh 5.39 A charge isolate metal sphere of iameter 0 cm has a potential of 8000V relative to V0 at infinity. alculate the energy ensity in the electric fiel near the surface of the sphere.

7 To compute the energy ensity in the fiel, we nee to fin the fiel. Since this charge is on a sphere, we can compute the fiel easily from the potential, assuming a point charge. q V 4rf0R q 4rf 0R V q E 4rf0R 4rf 0R V V 4rf0R R u E f0e f0 a V R k 0.3 J/m 3 (Note: The raius is 0.05m) 5.4 Given a 7.4pF air-fille capacitor, you are aske to convert it to a capacitor that can store up to 7.4 nj with a maximum potential ifference of 65 V. Which ielectric in Table 5- shoul you use to fill the gap in the capacitor if you o not allow for a margin of error? We nee to fin the require capacitance an then fin the ielectric constant that we nee to construct that capacitor from the air-fille capacitor that we have. We shoul then be able to ientify the material. U V U V # 0 - F l 0 l pF 7.4pF We nee to you Pyrex 5.48 Figure 5-48 shows a parallel-plate capacitor with a plate area A 5.56cm an place separation 5.56mm. The left half of the gap is fille with material of ielectric constant 7.00; the right half is fille with material of ielectric constant κ.00 What is the capacitance? You can think of this as two capacitors in parallel--each with half the area an its own ielectric constant.

8 A (A / ) κ A κ (A / ) parallel + (A / ) ( + κ ) A + κ (A / ) ( ) m m F 5.49 Figure 5-49 shows a parallel plate capacitor with a plate area A 7.89cm an plate separation 4.6mm. The top half of the gap is fille with material of ielectric constant.00; the bottom half is fille with material of ielectric constant κ.00 What is the capacitance? You can think of this as two capacitors in parallel--each with half the area an its own ielectric constant. A κ A A / κ A / + / series A + / κ A series / A ( + κ ) / A ( + κ κ ) A / ( κ + κ ) m m / F ( + ) 5.50 Figure 5-50 shows a parallel-plate capacitor of plate area A 0.5cm an plate separation 7.mm. The left half of the gap is fille with material of ielectric constant.0 ; the top of the right half is fille with material of ielectric constant κ 4.0 ; the

9 bottom of the right half is fille with material of ielectric constant κ What is the capacitance. This capacitor can be thought of as three capacitors as shown below 3 A / κ A / 3 κ 3 A /.37 0 F F F an 3 in series F F F an in parallel eq.37 0 F F F