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You are given two positively charged particles, of equal magnitude, separated by a distance, "d". What will happen to the force field between the two particles when "d" is doubled? A) remain constant B) decrease by a factor of two C) decrease by a factor of four D) nothing; no force field between particles

Respuesta :

osprey2015

force goes as 1/d^2 ... (2d)^2 => 4d^2 ...

C) decrease by a factor of four

Explanation :

There exists a force between charged particle given by Coulomb's law.

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

where,

k is the electrostatic constant

[tex]q_1,q_2[/tex] are charged particles

d is the distance between the charges.

When d is doubled (d' = 2d) , let F' be the force.

[tex]F'=k\dfrac{q_1q_2}{(2d)^2}[/tex]

[tex]F'=k\dfrac{q_1q_2}{4d^2}[/tex]

[tex]F'=\dfrac{1}{4}F[/tex]

So, the force decrease by a factor of four.

Hence, the correct option is (C) "decrease by a factor of four".

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