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Question-104572




Question Number 104572 by ajfour last updated on 22/Jul/20
Commented by ajfour last updated on 22/Jul/20
A point charge q with mass m  is released at a distance a from  a fixed uniformly charged rod  of length L, charge Q. Find the  speed acquired by point charge  q as it reaches an infinite  distance from the rod.
ApointchargeqwithmassmisreleasedatadistanceafromafixeduniformlychargedrodoflengthL,chargeQ.Findthespeedacquiredbypointchargeqasitreachesaninfinitedistancefromtherod.
Answered by OlafThorendsen last updated on 22/Jul/20
E = (Q/(2πε_0 )).(1/x).(1/(x+L))  F = qE = mx^(••)   ((qQ)/(2πε_0 )).(1/(x(x+L))) = mx^(••)   ((qQ)/(2πε_0 L)).(x^• /(x(x+L))) = mx^(••) x^•  (with v = x^• )  ((qQ)/(2πε_0 L^2 ))x^• [(1/x)−(1/(x+L))] = mx^(••) x^•   ((qQ)/(2πε_0 L^2 )).ln(x/(x+L)) = (1/2)mv^2 +C  t = 0, v = 0 ⇒ C = −((qQ)/(2πε_0 L^2 )).ln(a/(a+L))  (1/2)mv^2  = ((qQ)/(2πε_0 L^2 ))(ln(x/(x+L))−ln(a/(a+L)))  v_∞ ^2  = ((qQ)/(πε_0 mL^2 ))ln((a+L)/a)  ...may be
E=Q2πϵ0.1x.1x+LF=qE=mxqQ2πϵ0.1x(x+L)=mxqQ2πϵ0L.xx(x+L)=mxx(withv=x)qQ2πϵ0L2x[1x1x+L]=mxxqQ2πϵ0L2.lnxx+L=12mv2+Ct=0,v=0C=qQ2πϵ0L2.lnaa+L12mv2=qQ2πϵ0L2(lnxx+Llnaa+L)v2=qQπϵ0mL2lna+Lamaybe
Commented by ajfour last updated on 22/Jul/20
Dont you think Sir,  E=(Q/(4πε_0 ))(1/(x(x+L)))    ?
DontyouthinkSir,E=Q4πϵ01x(x+L)?

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