Question Number 38291 by Tinkutara last updated on 23/Jun/18
![](https://www.tinkutara.com/question/4477.png)
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](https://www.tinkutara.com/question/Q38293.png)
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18
![](https://www.tinkutara.com/question/4482.png)
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...](https://www.tinkutara.com/question/Q38342.png)
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](https://www.tinkutara.com/question/Q38349.png)
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.