12 HMK #3 Chp 24


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Unformatted text preview: 24-1CAPACITANCE AND DIELECTRICSNote: These solutions use Pfor the permittivity of free space. P= = 8.85E-12 F/m. 24.1. IDENTIFY:The capacitance depends on the geometry (area and plate separation) of the plates. SET UP:For a parallel-plate capacitor, Vab=Ed,E=Q!A, and C=QVab.EXECUTE:(a)Vab=Ed=4.00!106V/m( )2.50!10"3m( )=1.00!104V.(b)Solving for the area gives A=QE!=80."10#9C(4.00"106V/m)[8.854"10#12C2/(N$m2)]=2.26"10#3m2=22.6 cm2.(c)C=QVab=80.!10"9C1.00!104V=8.00!10"12F=8.00 pF.EVALUATE:The capacitance is reasonable for laboratory capacitors, but the area is rather large. 24.2. IDENTIFYand SET UP:ACd=P, QCV=and vEd=. (a)C = PA/d = P0 (0.00122m2) / (0.00328m) = 3.29 pF (b) 8124.3510C13.2 kV3.2910FQVC!!"==="(c) 3613.210 V4.0210 V/m0.00328 mVEd!===!EVALUATE:The electric field is uniform between the plates, at points that aren't close to the edges. 24.5. IDENTIFY:abQCV=. ACd=P. SET UP:When the capacitor is connected to the battery, 12.0 VabV=. EXECUTE:(a)64(10.010F)(12.0 V)1.2010C120 C abQCV!!=="="=(b)When dis doubled Cis halved, so Qis halved. 60 CQ=. (c) If ris doubled, Aincreases by a factor of 4. Cincreases by a factor of 4 and Qincreases by a factor of 4. 480 C.Q=EVALUATE:When the plates are moved apart, less charge on the plates is required to produce the same potential difference. With the separation of the plates constant, the electric field must remain constant to produce the same potential difference. The electric field depends on the surface charge density, !. To produce the same !, more charge is required when the area increases. 2424-2Chapter 2424.8. INCREASE:abQCV=. abVEd=. ACd=P. SET UP:We want 41.0010 N/CE=!when 100 VV=. EXECUTE:(a)2241.0010 V1.0010m1.00 cm1.0010 N/CabVdE!"==="=". 122321222(5.0010F)(1.0010m)5.6510m8.85410C /(N m )CdA!!!!""===""#P. 2Ar!=so 24.2410m4.24 cmAr!"==#=. (b)12210(5.0010F)(1.0010 V)5.0010C500 pCabQCV!!==""="=EVALUATE:ACd=P. We could have a larger d, along with a larger A, and still achieve the required Cwithout exceeding the maximum allowed E. 24.12. IDENTIFYand SET UP:Use the expression for C/Lderived in Example 24.4. Then use Eq.(24.1) to calculate Q.EXECUTE:(a)From Example 24.4, ()2ln/baCLrr!=P()()12221128.85410C / N m6.5710F/m66 pF/mln 3.5 mm/1.5 mmCL!""#$==#=(b) ()()11106.5710F/m2.8 m1.8410F.C!!="="()()103111.8410F35010V6.410C64 pCQCV!!!==""="=The conductor at higher potential has the positive charge, so there is +64 pC on the inner conductor and 64!pC on the outer conductor. EVALUATE:Cdepends only on the dimensions of the capacitor. Qand V are proportional. 24.16. IDENTIFY:The capacitors between band care in parallel. This combination is in series with the 15 pF capacitor. SET UP:Let 115 pFC=, 29.0 pFC=and 311 pFC=....
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This note was uploaded on 12/18/2011 for the course PHYS 2B 2b taught by Professor Hirsch during the Spring '10 term at UCSD.

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