Question

if the distance b/w two point chargers increased by 3% then calculate the % decrease in force b/w them

Answer 1

let the initial distance between the point charges is r

coulomb force , F α 1/r²      eq. 1
when distance is increased by 3% 

r1 = r + 0.03r = 1.03 r 

coulomb force ,F1 α 1/(1.03r)²      eq.  2

dividing eq.1 and 2 

F/F1 = (1.03r)²/r² = 1.06r²/r²  = 1.06

⇒F1 = F/1.06

decrease in force b/w charges 

= (F1 - F)/F  x 100  = [(F/1.06)  -  F]/F  x 100 = (F - 1.06 F/ 1.06 F) x 100

= (-0.06 F / 1.06 F) X 100 = -0.0566 x 100 =  5.66% 

- sign shows decrease in force 

hope,it helps 

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Answer 2

sol.
we know that,
force b/w charges =kQq/ r^2
so,
F= 1/r^2
therefore,
∆F/F= 2∆r/r
∆F/F×100=2∆r/r ×100
percentage change =2×3
=6
By 6%Force will be decreased if we increase r by 3%