Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with another whose

Question

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with another whose net charge is + 4Q. We move the +Q and +4Q charges to be 3 times as far apart as they were. What is the magnitude of the force on the +4Q charge ?

A. F

B. 4F

C. 4F/3

D. 4F/9

E. F/3

in progress 0
Tài Đức 1 week 2021-07-21T05:24:22+00:00 1 Answers 3 views 0

Answers ( )

    0
    2021-07-21T05:25:29+00:00

    Answer:

    F’= 4F/9

    Explanation:

    Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

    F=\dfrac{kQ^2}{r^2} …(1)

    Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

    F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}…..(2)

    Dividing equation (1) and (2), we get :

    \dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

    Hence, the correct option is (d) i.e. ” 4F/9″

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )