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{1}{4 Π ϵ_{0}} \times \frac{e}{d^{2}} e = c h a r g e o f e l e c t r o n

= 9 \times 10^{9} \times \frac{1 \times 1.6 \times 10^{- 19}}{{(10^{- 10})}^{2}} = 14.4 \times 10^{9 - 19 + 20}

= 1.44 \times 10^{11} f o r e l e c t r o n

s a m e v a l u e f o r p r o t o n B u t o p p o s i t e d i r e c t i o n

s o n e t e l e c t r i c f i e l d i s z e r o

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...

p l s c h e c k n o w i f o r g o t t o v e c t o r a d d i t i o n

o f p r o t o n a n d e l e c t r o n \dots

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

i n s u b a t o m i c c a s e p r o t o n a l s o c o m e s t o p l a y . .

b e c a u s e s e p a r a t i o n i n A^{0} \dots p l s s e a r c h i n

w i k i p e d i a o r p o s t t h e s a m e q u e s t i o n i n Q U O R A

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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