Two like charges exert a force of repulsion of 3.5 N on each other. What will be the force of the distance between the charges increased to five times the original value? 

Asked by tippanjanhavi5320.12sdatl | 8th May, 2020, 01:31: PM

Expert Answer:

By Coulomb's law, 
 
F space equals space fraction numerator k q subscript 1 q subscript 2 over denominator r squared end fraction
I n t i a l l l y space t h e space f o r c e space o f space r e p u l s i o n space b e t w e e n space t w o space c h a r g e s space i s space 3.5 space N space
T h u s comma space

F subscript 1 space equals space fraction numerator k q subscript 1 q subscript 2 over denominator r squared end fraction space
rightwards double arrow 3.5 space equals fraction numerator k q subscript 1 q subscript 2 over denominator r squared end fraction space space.... space left parenthesis 1 right parenthesis

L e t space F subscript 2 space b e space t h e space f o r c e space b e t w e e n space t h e space c h a r g e s space w h e n space d i s tan c e space b e t w e e n space t h e m space i s space i n c r e a s e d space 5 space t i m e s space t h e space o r i g i n a l space v a l u e space

F subscript 2 space equals space space fraction numerator k q subscript 1 q subscript 2 over denominator open parentheses 5 r close parentheses squared end fraction space equals space space fraction numerator k q subscript 1 q subscript 2 over denominator 25 r squared end fraction space... space left parenthesis 2 right parenthesis space

T a k i n g space r a t i o space o f space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis comma space w e space g e t comma space
fraction numerator 3.5 over denominator F subscript 2 end fraction space equals space fraction numerator fraction numerator k q subscript 1 q subscript 2 over denominator r squared end fraction over denominator fraction numerator k q subscript 1 q subscript 2 over denominator 25 r squared end fraction end fraction space equals space fraction numerator k q subscript 1 q subscript 2 over denominator r squared end fraction cross times fraction numerator 25 r squared over denominator k q subscript 1 q subscript 2 end fraction space

H e n c e comma space

fraction numerator 3.5 over denominator F subscript 2 end fraction space equals space 25
rightwards double arrow F subscript 2 space equals space fraction numerator 3.5 over denominator 25 end fraction space equals 0.14 space N space
Thus, the force when distance between the charges is increased to five times = 0.14 N 

Answered by Shiwani Sawant | 8th May, 2020, 02:56: PM