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Which of the following changes would double the force between two charged particles? O A. Decreasing the distance between the particles by a factor of 2 B. Increasing the distance between the particles by a factor of 2 O C. Doubling the amount of charge on one of the particles O D. Doubling the amount of charge on each particle

Sagot :

Let's check

[tex]\\ \rm\dashrightarrow F=\dfrac{k}{q_1q_2}{r^2}[/tex]

[tex]\\ \rm\dashrightarrow F\propto Q[/tex]

[tex]\\ \rm\dashrightarrow F\propto \dfrac{1}{r^2}[/tex]

So

Option A and C can be used

SmaгtieGurLidn

Answer:

Decreasing the distance between the particles by a factor of 2

Explanation:

To double the force between the two charge particles, the distance between the particles should be reduced by a factor of two.

According to coulombs law "force is directly proportional to the potential between the two charges and inversely proportional to the square of their distances".

       [tex]\sf{F  = \dfrac{kq_{1} q_{2} }{r^{2} }}[/tex]

F is the electric force

k is the coulomb constant

q is the charge

r is the distance or separation

As the separation between the charges is reduced the force increases and vice versa.

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