Question

Please help me with this..Two charged objects have a repulsive force of 0.14 N. If the charge of one of the objects is doubled , and the distance separating the objects is doubled , then what is the new force ?

Answer 1

According to Coulomb's law force between 2 charged objects is

`F=(k_e* q_1* q_2)/(r^2)`

Where q1 and q2 are the charges, r is the distance between them and

`k_e\text{ is Coulomb's constant }`
`k_e=(1)/(4\pi\epsilon_0)=8.98*10^9\text{ kg}\cdot m^3\cdot s^(-2)\cdot C^(-2)`

Now,

`F_1=(k_e* q_1* q_2)/(r^2)=0.14`

Let say q2 changes its value to

`2q_2_{}`

and r changes to 2r

Now the new force will be

`F_2=(k_e* q_1*2q_2)/((2r)^2)=(k_e* q_1* q_2)/(2r^2)`

Now dividing the two values of force

`(F_1)/(F_2)=(k_eq_1q_2*2r^2)/(k_eq_1q_2* r^2)`

ON further simplifying

`(F_1)/(F_2)=2`

Or

`F_2=(F_1)/(2)=(0.14)/(2)=0.07\text{ N}`

i.e. new force is 0.07N