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A charge is distributeduniformly overa ringof radius'a'Obtain anexpressionfor theelectricintensityE at apointon theaxisof thering . Henceshow thatforpointa lergedistancefrom theringitbehaves likea pointcharge . |
Answer» Solution : Suppose that theringis placed withits planeperpendicularto thex-axisas shown in Fig. Consider asmallelementdl of the ring .  Asthe totalchargeq isuniformlydistributedthe charge dq on the elementdl isdq `=(Q)/(2pia) .dl` `:.` Themagnitudeof thefieldd `bar(E)` produced bytheelementdlat thefieldpoint P is `bar(DE) =k .(dq)/(r^(2)) =(kq)/(2pia) .(dl)/(r^(2))` Thefield `dbar(E)` has twocomponents (a) the axialcomponent dEcos `theta` , and (b)the perpendicularcomponentdEsin `theta` Sincethe perpendicularcomponentsof any twodiametricallyoppositeelements are equalandoppositethey allcancelout inpairs.Onlythe axialcomponents will addup toproducethe resultantfield `bar(E)` at pointP whichis givenby `E = overset(2pia)underset(0)(int) dE " cos" theta""[:' "onlythe axialcomponentscontributetowards" E]` `=overset(2pia)underset(0)(int) (kq)/(2pia) ,(dl)/(r^(2)).(X)/(r ) =(kpx)/(2pia) .(1)/(r^(3)) overset(2pia)underset(0)(int) dl` `=(kpx)/(2pia).(1)/(r^(3)) [l]_(0)^(2pia) =(kpx)/(2pia).(1)/((x^(2)+a^(2))^(3//2)).2pia ""[:' r^(2) =x^(2) +a^(2)]` `"or"""E =(kpx)/((x^(2) +a^(2))^(3//2))=(1)/(4piepsilon_(0)).(qx)/((x^(2) +a^(2))^(3//2))` LTBGT `" if" ""x gt gt a` `x^(2) +a^(2) ~~x^(2)` `E =(1)/(4pi epsilon_(0)) .(qx)/(x^(3)) rArr E=(1)/(4piepsilon_(0)).(q)/(x^(2))` i.e., Whenthe pointis FARAWAY from thecentrecharged ringacts likea pointcharge . |
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