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Question-38291




Question Number 38291 by Tinkutara last updated on 23/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18
E=(1/(4Πε_0 ))×(e/d^2 )   e=charge of electron  =9×10^9 ×((1×1.6×10^(−19) )/((10^(−10) )^2 ))=14.4×10^(9−19+20)   =1.44×10^(11)  for electron  same value for proton But opposite direction  so net electric field is zero
E=14Πϵ0×ed2e=chargeofelectron=9×109×1×1.6×1019(1010)2=14.4×10919+20=1.44×1011forelectronsamevalueforprotonButoppositedirectionsonetelectricfieldiszero
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18
Commented by Tinkutara last updated on 24/Jun/18
Answer is wrong, Sir.
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18
plscheck now i forgot to vector addition  of proton and electron...
plschecknowiforgottovectoradditionofprotonandelectron
Commented by Tinkutara last updated on 24/Jun/18
Answer given is b.
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18
in sub atomic case proton also comes to play..  because separation in A^0 ...pls search in   wikipedia or post the same question in QUORA
insubatomiccaseprotonalsocomestoplay..becauseseparationinA0plssearchinwikipediaorpostthesamequestioninQUORA
Commented by prakash jain last updated on 24/Jun/18
distance between alpha particle in electron is 1 A then what is the distance between alpha particle and proton? I think proton will give negilgible field.

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