Two point charges, initially 2 cm apart, are moved to a distance of 10 cm apart. by what factor does the resulting electric force
between them change?
a 25
b.5
c.1/5
d.1/25

1/25

Explanation:

Coulombs law states that the force of attraction between two charges q and Q is directly proportional to the product of the charges and inversely proportional to the square of the distance r between them.

It is expressed as;.

F = kqQ/r²

If two point charges, initially 2 cm apart, the force between them will be;

F = kqQ/2²

F = kqQ/4 ... (1)

If the charges are now moved to a distance of 10cm apart, the force between them will then become;

F2 = kqQ/10²

F2 = kqQ/100.. (2)

Dividing equation 2 by 1;

F2/F = (kqQ/4)/kqQ/10)

F2/F = kqQ/100 × 4/kqQ

F2/F = 1/25

F2 = (1/25)F

The resulting electric force

between the changes by 1/25 since that serves as the constant of proportionality between the two forces

he . if i may ask what i your question exactly?

explanation: