Respuesta :
As we know that force between two charges is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now we know that the distance between two charges is tripled and one of the charge is quadrupled
so the new force is given as
[tex]F' = \frac{kq_1(4q_2)}{(3r)^2}[/tex]
now we have
[tex]F' = \frac{4kq_1q_2}{9r^2}[/tex]
so we have to take ratio of two equations
[tex]\frac{F}{F'} = \frac{9}{4}[/tex]
on rearranging above equation we have
[tex]F' = \frac{4}{9}F[/tex]
so fourth option is correct answer
Answer
Correct answer is 4(F′ =(4/9)F)
Explanation
According to coulombs law,
for two charges [tex]q_{1}[/tex] and [tex]q_{2}[/tex] which are separated by a distance d and exert a force F on each other which is given by following equation
[tex]F=\frac{9\times 10^9\times q_{1}\times q_{2}}{d^2}[/tex]............(1)
If if [tex]q_{1}[/tex] is quadrupled and d is tripled then force F' will be given as
[tex]F'=\frac{9\times 10^9\times 4q_{1}\times q_{2}}{(3d)^2}[/tex]
[tex]F'=\frac{9\times 10^9\times 4q_{1}\times q_{2}}{9d^2}[/tex]
[tex]F'=\frac{4}{9}\times\frac{9\times 10^9\times q_{1}\times q_{2}}{d^2}[/tex].....(2)
therefore, from equation (1) and equation(2)
[tex]F'=\frac{4F}{9}[/tex]