Question Number 38291 by Tinkutara last updated on 23/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18
$${E}=\frac{\mathrm{1}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} }×\frac{{e}}{{d}^{\mathrm{2}} }\:\:\:{e}={charge}\:{of}\:{electron} \\ $$$$=\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{1}×\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} }{\left(\mathrm{10}^{−\mathrm{10}} \right)^{\mathrm{2}} }=\mathrm{14}.\mathrm{4}×\mathrm{10}^{\mathrm{9}−\mathrm{19}+\mathrm{20}} \\ $$$$=\mathrm{1}.\mathrm{44}×\mathrm{10}^{\mathrm{11}} \:{for}\:{electron} \\ $$$${same}\:{value}\:{for}\:{proton}\:{But}\:{opposite}\:{direction} \\ $$$${so}\:{net}\:{electric}\:{field}\:{is}\:{zero} \\ $$
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
$${plscheck}\:{now}\:{i}\:{forgot}\:{to}\:{vector}\:{addition} \\ $$$${of}\:{proton}\:{and}\:{electron}... \\ $$
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
$${in}\:{sub}\:{atomic}\:{case}\:{proton}\:{also}\:{comes}\:{to}\:{play}.. \\ $$$${because}\:{separation}\:{in}\:{A}^{\mathrm{0}} ...{pls}\:{search}\:{in}\: \\ $$$${wikipedia}\:{or}\:{post}\:{the}\:{same}\:{question}\:{in}\:{QUORA} \\ $$
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.