If the force between two charged objects when you triple the magnitude of both charges. The Force will become 9 times the initial force.
According to coulombs law force between two charges is given by
[tex]F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}[/tex]
Here, R is the distance between both the charges Q and q.
Let the force on the charges be F1
The distance of separation = r
The magnitudes of the charges q1 and q2
K = Coulumb's constant
[tex]F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}[/tex]
Let the force on the charges be F
The distance of separation = r
The magnitudes of the charges 3q1 and 3q2
K = Coulumb's constant
So,
[tex]F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}[/tex]
[tex]F = \dfrac{1}{4\pi \epsilon }\dfrac{3Q\times 3q}{r^2}\\\\F = \dfrac{1}{4\pi \epsilon }\dfrac{9Qq}{r^2}\\\\F =9 \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}\\\\F = 9F_1[/tex]
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