Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 n. what would be the force between two charges q1 and= 2q and q2=3q, that are distance d apart?

Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?

The force = 38.4 N

Explanation:

From coulombs law,

F = kq₁q₂/r² Equation 1

Where F = Force of attraction or repulsion between the charges, q₁ and q₂ = first and second charge respectively, r = distance between the charges, k = constant of proportionality.

When, F = 6.4 N, r = d m.

6.4 = kq₁q₂/d² Equation 1

When q₁' = 2q₁, q₂' = 3q₂, r = d cm

F = k(2q₁)(3q₂)/d²

F = 6kq₁q₂/d² Equation 2

Dividing Equation 1 by equation 2

6.4/F = kq₁q₂/d²/(6kq₁q₂/d²)

6.4/F = 1/6

F = 6.4×6

F = 38.4 N.

Thus the force = 38.4 N

it becomes positively charged by charging it

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s