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NEET Class neet Answered

the force between two electrons when placed in air is equal to 0.5 times the weight of an electrons find the distance between two electrons (mass of electrons=9.1*10to the power of -31  
Asked by kattasiddharth06 | 24 Aug, 2019, 09:25: AM
answered-by-expert Expert Answer
begin mathsize 14px style fraction numerator e squared over denominator 4 pi epsilon subscript o space r squared end fraction space equals space 0.5 space m subscript e g space space
r space equals space square root of fraction numerator 2 pi epsilon subscript o space m subscript e g over denominator e squared end fraction end root space equals space square root of fraction numerator 2 pi cross times 8.85 cross times 10 to the power of negative 12 end exponent cross times 9.1 cross times 10 to the power of negative 31 end exponent cross times 9.8 over denominator 1.6 cross times 1.6 cross times 10 to the power of negative 38 end exponent end fraction end root space equals space 0.139 space m end style
In the above calculation
 
e = electronic charge = 1.6×10-19 C
me = mass of electron = 9.1×10-31 kg
g = acceleration due to gravity = 9.8 m/s2
εo = permitivity of air = 8.85×10-12 F/m
r = distance between electron in m
Answered by Thiyagarajan K | 24 Aug, 2019, 10:31: AM
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