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    Question

    When the distance between two charged particles is halved, then the force between them becomes :

    A

    One-fourth

    B

    Half

    C

    Double

    D

    Four times

    Medium Open in App Solution Verified by Toppr

    Correct option is D)

    Earlier force between two particles is

    F= r 2 kq 1 ​ q 2 ​ ​

    When distance is halved force becomes

    F ′ = (r/2) 2 kq 1 ​ q 2 ​ ​ =4 r 2 kq 1 ​ q 2 ​ ​ =4F

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    If the distance between two charges is halved, what will happen the force between the charges?

    4 times the initial From Coloumb's law of electrostatic force we know, F = (k q_1 q_2)/r^2 (where, k is a constant, q_1 and q_2 are two charges separated by distance r) So,if the distance between the charges is halved, with no changes of the charges,new distance becomes r/2 So,if now force acting between the same charges is F" Then F'= (k q_1 q_2)/(r/2)^2 = 4 (k q_1 q_2)/r^2 = 4 F So,that means, force will be 4 times of the initial value

    If the distance between two charges is halved, what will happen the force between the charges?

    Physics

    2 Answers

    Jane Feb 26, 2018 4 times the initial

    Explanation:

    From Coloumb's law of electrostatic force we know,

    F = k q 1 q 2 r 2 (where, k is a constant, q 1 and q 2

    are two charges separated by distance

    r )

    So,if the distance between the charges is halved, with no changes of the charges,new distance becomes

    r 2

    So,if now force acting between the same charges is

    F Then F ' = k q 1 q 2 ( r 2 ) 2 = 4 k q 1 q 2 r 2 = 4 F

    So,that means, force will be

    4

    times of the initial value

    Answer link Shwetank Mauria Feb 26, 2018

    The force between the two charges is quadrupled.

    Explanation:

    The force between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Hence, if distance between charges is halved (charges remaining kept constant),

    the force between the two charges is quadrupled.

    Answer link

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    स्रोत : socratic.org

    When the distance between two charged particles is halved, the force between them will become

    r(2)=r(1)/2,F(2)=? Fprop1/r^(2) therefore" "F(2)/F(1)=(r(1)/r(2))^(2)=(r(1)/(r(1)/2))^(2)=4 F(2)=4F(1)="four times"

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    When the distance between two charged particles is halved, the force between them will become

    Updated On: 27-06-2022

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    Text Solution Open Answer in App A one-third B one-half C four times D five times Answer

    The correct Answer is C

    Solution r 2 = r 1 2 , F 2 =? r2=r12,F2=? F∝ 1 r 2 F∝1r2 ∴ F 2 F 1 = ( r 1 r 2 ) 2 = ( r 1 r 1 2 ) 2 =4

    ∴ F2F1=(r1r2)2=(r1r12)2=4

    F 2 =4 F 1 =four times F2=4F1=four times Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6

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