24-1CAPACITANCE AND DIELECTRICSNote: These solutions use P0for the permittivity of free space. P0= ε0= 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!0A, 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!0=80.0"10#9C(4.00"106V/m)[8.854"10#12C2/(N$m2)]=2.26"10#3m2=22.6 cm2.(c)C=QVab=80.0!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:0ACd=P, QCV=and VEd=. (a)C = P0A/d = P0 (0.00122m2) / (0.00328m) = 3.29 pF (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. 24.5. 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, !. To produce the same !, more charge is required when the area increases. 24
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