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

$F=\frac{k_e\times q_1\times q_2}{r^2}$

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

$k_e\text{ is Coulomb's constant }$$k_e=\frac{1}{4\pi\epsilon_0}=8.98\times10^9\text{ kg}\cdot m^3\cdot s^{-2}\cdot C^{-2}$

Now,

$F_1=\frac{k_e\times q_1\times q_2}{r^2}=0.14$

Let say q2 changes its value to

$2q_2_{}$

and r changes to 2r

Now the new force will be

$F_2=\frac{k_e\times q_1\times2q_2}{(2r)^2}=\frac{k_e\times q_1\times q_2}{2r^2}$

Now dividing the two values of force

$\frac{F_1}{F_2}=\frac{k_eq_1q_2\times2r^2}{k_eq_1q_2\times r^2}$

ON further simplifying

$\frac{F_1}{F_2}=2$

Or

$F_2=\frac{F_1}{2}=\frac{0.14}{2}=0.07\text{ N}$

i.e. new force is 0.07N