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Two charges attract each other with a force of 1.5 N. What will be the force if the distance between them is reduced to one-ninth of its original value?

Respuesta :

BHmohammad
F=k×q1×q2/r^2

if r becomes 1/9 r

F= 1.5/(1/9^2)
F= 1.5 ×9^2
F=121.5

Explanation:

The force between two charges is, F = 1.5 N when they are r distance apart such that,

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

[tex]1.5=k\dfrac{q_1q_2}{r^2}[/tex].......(1)

We need to find the force F' if the distance between them is reduced to one-ninth of its original value, r' = r/9

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

[tex]F'=k\dfrac{q_1q_2}{(r/9)^2}[/tex]..........(2)

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

[tex]\dfrac{1.5}{F'}=\dfrac{1}{81}[/tex]

F' = 121.5  N

So, the new force is 121.5  N when the distance between them is reduced to one-ninth of its original value. Hence, this is the required solution.        

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