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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
$${E}=\frac{\mathrm{1}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} }×\frac{{e}}{{d}^{\mathrm{2}} }\:\:\:{e}={charge}\:{of}\:{electron} \\ $$$$=\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{1}×\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} }{\left(\mathrm{10}^{−\mathrm{10}} \right)^{\mathrm{2}} }=\mathrm{14}.\mathrm{4}×\mathrm{10}^{\mathrm{9}−\mathrm{19}+\mathrm{20}} \\ $$$$=\mathrm{1}.\mathrm{44}×\mathrm{10}^{\mathrm{11}} \:{for}\:{electron} \\ $$$${same}\:{value}\:{for}\:{proton}\:{But}\:{opposite}\:{direction} \\ $$$${so}\:{net}\:{electric}\:{field}\:{is}\:{zero} \\ $$
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...
$${plscheck}\:{now}\:{i}\:{forgot}\:{to}\:{vector}\:{addition} \\ $$$${of}\:{proton}\:{and}\:{electron}… \\ $$
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
$${in}\:{sub}\:{atomic}\:{case}\:{proton}\:{also}\:{comes}\:{to}\:{play}.. \\ $$$${because}\:{separation}\:{in}\:{A}^{\mathrm{0}} …{pls}\:{search}\:{in}\: \\ $$$${wikipedia}\:{or}\:{post}\:{the}\:{same}\:{question}\:{in}\:{QUORA} \\ $$
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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