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CBSE Class 12-science Answered

A point charge qa = + 100 c is placed at point a (1, 0, 2) m and another point charge qb = +200c isplaced at point b (4, 4, 2) m. Find :(i) magnitude of electrostatic interaction force acting between them(ii) find a f(force on a due to
Asked by raghavsumit9690 | 10 Jul, 2021, 01:30: PM
Expert Answer
Electrostatic interaction force F between charges is given as
 
begin mathsize 14px style F space equals space K space cross times fraction numerator q subscript a space q subscript b over denominator d squared end fraction end style
Where K = 1/(4πεo ) = 9 × 109 N m2 C-2  is Coulomb's constant  and εo is permittivity of free space .
 
Distance between charges is given as d in above expression
 
d = begin mathsize 14px style square root of open parentheses 1 minus 4 close parentheses squared space plus space open parentheses 0 minus 4 close parentheses squared space plus space left parenthesis 2 minus 2 right parenthesis squared end root space equals space 5 space m end style
F = ( 9 × 109 × 100 × 200 ) / 25  = 7.2 × 1012 N
 
----------------------------------
 
Force on charge qa due to charge qb = 7.2 × 1012 N
 


Answered by Thiyagarajan K | 10 Jul, 2021, 03:33: PM

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