Answer:
The distance was decreased 5 times. Thus, the factor of decrease is 1/5.
Step-by-step explanation:
The electric force (F) between 2 charges can be described through Coulomb's law.
[tex]F_{1}=k\frac{q_{1}q_{2}}{d_{1}^{2} }[/tex]
where,
K: Coulomb's constant
q: charges
d: distance
If F₂ = 25 F₁,
[tex]F_{2}=k\frac{q_{1}q_{2}}{d_{2}^{2} }\\25F_{1}=k\frac{q_{1}q_{2}}{d_{2}^{2} }\\25k\frac{q_{1}q_{2}}{d_{1}^{2} }=k\frac{q_{1}q_{2}}{d_{2}^{2} }\\\frac{d_{2}^{2} }{d_{1}^{2} } =1/25\\\frac{d_{2} }{d_{1} } =1/5[/tex]
The distance was decreased 5 times.
