Question

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

Answer 1

Correct answer is 4(F′ =(4/9)F)

Step-by-step explanation

According to coulombs law,

for two charges
`q_(1)` and
`q_(2)` which are separated by a distance d and exert a force F on each other which is given by following equation

`F=(9* 10^9* q_(1)* q_(2))/(d^2)`............(1)

If if
`q_(1)` is quadrupled and d is tripled then force F' will be given as

`F'=(9* 10^9* 4q_(1)* q_(2))/((3d)^2)`

`F'=(9* 10^9* 4q_(1)* q_(2))/(9d^2)`

`F'=(4)/(9)*(9* 10^9* q_(1)* q_(2))/(d^2)`.....(2)

therefore, from equation (1) and equation(2)

`F'=(4F)/(9)`

Answer 2

As we know that force between two charges is given as

`F = (kq_1q_2)/(r^2)`

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

`F' = (kq_1(4q_2))/((3r)^2)`

now we have

`F' = (4kq_1q_2)/(9r^2)`

so we have to take ratio of two equations

`(F)/(F') = (9)/(4)`

on rearranging above equation we have

`F' = (4)/(9)F`

so fourth option is correct answer