24-1CAPACITANCE AND DIELECTRICS24.1. IDENTIFY:abQCV=SET UP:61 F10Fμ−=EXECUTE:64(7.2810F)(25.0 V)1.8210C182 CabQCVμ−−==×=×=EVALUATE:One plate has charge Q+and the other has charge Q−. 24.2. IDENTIFYand SET UP:0ACd=P, QCV=and VEd=. (a)2000.00122 m3.29 pF0.00328 mACd===PP(b) 8124.3510C13.2 kV3.2910FQVC−−×===×(c) 3613.210V4.0210V/m0.00328 mVEd×===×EVALUATE:The electric field is uniform between the plates, at points that aren't close to the edges. 24.3. IDENTIFYand SET UP:It is a parallel-plate air capacitor, so we can apply the equations of Sections 24.1. EXECUTE:(a)6120.14810Cso 604 V24510FababQQCVVC−−×====×(b) 0soACd=P()()1233221222024510F0.32810m9.0810m90.8 cm8.85410C / N mCdA−−−−××===×=×⋅P(c) 63604 Vso 1.8410V/m0.32810mababVVEdEd−====××(d) 0soEσ=P()()612225201.8410V/m8.85410C/ Nm1.6310C/mEσ−−==××⋅=×PEVALUATE:We could also calculate σdirectly as Q/A. 652320.14810C1.6310C/m ,9.0810mQAσ−−−×===××which checks. 24.4. IDENTIFY:0ACd=Pwhen there is air between the plates. SET UP:22(3.010m)A−=×is the area of each plate. EXECUTE:1222123(8.85410F/m)(3.010m)1.5910F1.59 pF5.010mC−−−−××==×=×EVALUATE:Cincreases when Aincreases and C increases when ddecreases. 24.5. IDENTIFY:abQCV=. 0ACd=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 abQCVμ−−==×=×=(b)When dis doubled Cis halved, so Qis halved. 60 CQμ=. (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.Qμ=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, σ. To produce the same σ, more charge is required when the area increases. 24
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