Chapter 2313P23.33Due to symmetryEdEyy==z0 , and Ekdqrxezzsinsinθ2wheredqdsrdλλ,so that,Ekrdkrkrxeee−=zλθθπsincos002afwhere=qLand rL=.FIG. P23.33Thus,EkqLxe×⋅×−22 8 99 107 50 1001402962...Nm CCm22ejej.Solving,Ex=×216 107. NC.Since the rod has a negative charge, Eii=−×= −2167.±.±NCMNC .P23.34(a)We define x=0 at the point where we are to find the field. One ring, with thickness dx, hascharge Qdxhand produces, at the chosen point, a fielddkxxRQdxhe=+32±.The total field isEEiiEii+=+=+−=+−++LNMMMOQPPPz+−=+−=+dkQxdxhxRkQhxdxhhdRdh Reddhexdeeall charge122222211bg±±±±(b)Think of the cylinder as a stack of disks, each with thickness dx, charge Qdxh, and charge-per-area σ=QdxRh2. One disk produces a fielddkQdxxe+FHGGIKJJ212±.So,i−+FHGGIKJJ=+dxeall charge
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .