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(a) Edectric field{:[E=10000N//C],[=>(a)/(atz_(0)r^(2))=10000],[" or "q=10^(4)xx4pit_(0)(0.250)^(2)],[=6.954 xx10^(-8C)C],[:.a=69.54 nC],[(1n{:=10^(-9))]:}(b).2. quad F=(q_(1)q_(2))/(4pit_(0)r^(2))=q_(2)E_(1)E_(1)= field dueto q_(1) charge.q_(2)=-1.75 xx10^(-6)cfree exerts upword.{:[F=2xx10^(5).N],[" 2 "50E_(1)=quad f//a_(2)],[=-(2xx10^(-5))/(.1.75 xx10^(-6))N//c],[=-11.4285N//C]:}if we take q_(1) ar origin thenE_(1) andentromsing in entering magnitioe is |E_(1)|=11.4285N//CFree is up word So F=2xx10^(-5) hat(j)35.F=(q_(1)q_(2))/(4pit_(0)r^(2))So f prop(1)/(r^(2))So if r becom 3r then f^(')alpha(1)/((3r)^(2)){:[2(1)/(ar^(2))],[:.quad(F^('))/(F)=(r^(2))/(9r^(2))=(1)/(9)],[:.F^(')=(F)/(9)]:}So force will decrease by factor ? or Frace will be (1)/(n) th of the riginal one{:[⨆_("exst ")^("North ")longrightarrow^(E)],[" Force "f=9E],[=(3.5 xx10^(-6)xx250)],[=8.75 xx10^(-4)" trwards "],[=8" east. "]:}(a) E=E_(1)+E_(2) Non quadE_(1)=(q_(1))/(4pit_(0)r_(1)^(2))=(1.0 xx10^(-6))/(4pit_(0)(2+1)^(3)) E_(2) > (q_(2))/(4pit_(0)r_(2)^(2))=(-3.0 xx10^(6))/(4pit_(0)(1)^(3))NonSo{:[E_(1)=332.87N//C],[E_(2)=-26962.65N//c]:}So resultant electric fieldE=-26629.78N//e[["-ve sign "],[" indicates the "],[E]:}the directoon in towards the charge -3. ... See the full answer