Question

If the distance between two particles is halved, how is the force of attraction affected?? plz explain in easy words, according to class 9..

Answer 1

Let us assume that we are talking about graviational force of attraction.

           F = G M1 M2 / d²

F is the force that one object pulls the other towards itself.

Here M1 = mass of 1st particle      M2 = mass of second particle
       d = distance between the two particles.
        G = 6.67 * 10^-11   units        [exponential -11]

This force is very less for small masses and significant for planets like Earth, Sun, moon etc.

Now, if  d becomes d / 2, substitute it in the equation above, to get

         F = G M1 M2 / (d/2)²    =    G M1 M2 / ( d² / 4 )    = GM1 M2 *4 / d²
             =  4 *  [G M1 M2 / d²]

So the Force has become 4 times the previous value.

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If we are talking about force between electric charges, then also the same result will be observed.  The new value of force will be 4 times the original value.

Answer 2

As u have not mentioned so m taking gravitational forces of attraction as u said u r in class 9
F=G(m1m2/r²)--------(1)
where,F is the force
m1=1st mass
m2=2nd mass
G=Gravitational constant
r is distance between centre of masses(m1 and m2).
If r becomes(r/2)half,then
putting this in the above equation,-------------(1)
F=G{m1m2/(r/2)²}
or,F=G{m1m2/(r²/4)}
or,4F=G{m1m2/r²}
Therefore force will become 4 times greater than the previous force.