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SIZIF [17.4K]
3 years ago
8

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 an

other 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
Physics
1 answer:
Alona [7]3 years ago
7 0

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"

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