Question

Two point charges are originally placed a certain distance apart, and the force between them is measured. The two charges are then brought closer together so that the force between them increases by a factor of 25. By what factor was their separation decreased?

Answer 1

The distance was decreased 5 times. Thus, the factor of decrease is 1/5.

Explanation:

The electric force (F) between 2 charges can be described through Coulomb's law.

`F_(1)=k(q_(1)q_(2))/(d_(1)^(2) )`

where,

K: Coulomb's constant

q: charges

d: distance

If F₂ = 25 F₁,

`F_(2)=k(q_(1)q_(2))/(d_(2)^(2) )\\25F_(1)=k(q_(1)q_(2))/(d_(2)^(2) )\\25k(q_(1)q_(2))/(d_(1)^(2) )=k(q_(1)q_(2))/(d_(2)^(2) )\\(d_(2)^(2) )/(d_(1)^(2) ) =1/25\\(d_(2) )/(d_(1) ) =1/5`

The distance was decreased 5 times.