Two charges q1 and q2 are separated by a distance d and exert a force F on each other.

What new force F′ would exist if q1 is quadrupled and d is tripled?
1. F′ = (9/4) F
2. None of these
3. F′ = 0
4. F′ =(4/9)F
5. F′ =(9/81)F
6. F′ =(81/16)F
7. F′ =(16/81)F
8. F′ =(81/9)F

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]