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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

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 =
F = ( 9 × 109 × 100 × 200 ) / 25  = 7.2 × 1012 N

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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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