Midterm 2 [F2015] [Solutions] (PHYS 301) - PHYS 301 Midterm...

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PHYS 301 Midterm II, 2015Joshua Smith, ID: 30890131November 3, 2016Problem 1:Two isolated square parallel conducting plates of sideLand separationdarecharged withQon the upper plate and-Qon the lower plate (see the figure).Two dielectric slabs, each with thicknessdand areaL22 are inserted betweenthe plates (side by side). The permitivities of the dielectric slabs are1and2.Assumed<<L.(a)FindDeverywhere between the plates. (Hint. Consider whether thesurface charge density on the platesσis constant.)(b)FindEeverywhere between the plates.(c)Calculate the bound surface charge densitiesσbon the dielectric surfacestouching the plates and between the plates.(d)Find the capacitance.Problem 1Solution:(a)In order to determineD, we use Gauss’ Law, which states thatDdA=QfD=σfwhereQfis the free charge andσfis the free charge density.By Gauss’ Law, we know that at in the plates,D=0, since there is noenclosed free charge. Additionally, we find thatD=σfeverywhere betweenthe two conducting plates.1
However, we cannot stop there. We know that the two dielectric slabs have thesame potential difference across them since the net charge drop across them isthe same. Since the two slabs have different permitivities, each slab will have adifferent electric fieldEinside, despite having the same potential differenceδVacross them. As such, we must say thatD=σ1if 0x<L2,0zdσ2ifL2xL,0zdwhereσ1is the free charge density above the first dielectric andσ2is the freecharge density above the second dielectric.We want to have these surface charge densities in terms of the total chargeQ.We know thatσ1L22+σ2L22=Qf=Qσ1+σ2=2QL2and we know that the electric fieldEis uniform throughout the twodielectrics. Since we know thatD=0E+P=0(1+χ)E=0rE=Ewe can say thatσ11=σ22σ2=σ121If we substitute this back in, we find thatσ11+21=2QL2σ1=2QL21+21σ1=2Q1L2(1+2)Since we knowσ1, we find thatσ2=σ121=2Q12L21(1+2)=2Q2L2(1+2)Therefore, we find thatD=2Q1L2(1+2)if 0x<L2,0zd2Q2L2(1+2)ifL2xL,0zdThis makes sense, asDdepends on the dielectric slab and is uniform betweenthe plates.(b)In order to calculate the electric fieldE, we re-useD=E. As such, wecan simply divide the electric displacementDby the permitivity of therespective dielectric material. We find thatE=σ11if 0x<L2,0<z<dσ22ifL2xL,0<z<d2
If we substitute our known values for the surface charge densities in, we findthatE=2QL2(1+2)

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