Question

If the distance between two particles is increased by 2%,then the force of attraction between them will decrease by what %?

Answer 1

we know, gravitational force or Coulomb's force both is inversely proportional to square of distance between particles .
e.g., $F\propto\frac{1}{r^2}$

so, F = K/r² , K is proportionality constant.

differentiate both sides with respect to r

dF/dr = -2K/r³

dF = -2Kdr/r³

dF/F = -2Kdr/r³F

dF/F = -2Kdr/r³ × (K/r²) = -2dr/r

To find maximum error ,
∆F/F = 2 × ∆r/r

so, % error in F = 2 × % error in r

given, % error in r = 2%
so, % error in F = 2 × 2 = 4%

Answer 2

Answer:

The answer is given below