Question

The electric force between two charged objects is 5.2 x 10-4 N when the objects are 3.11 x 10-1 m apart. What is the electric force between the same objects if the distance is changed to 4.04 x 10-1 m?

Answer 1

$F_1=5.2\times 10^-4 N \\ r_1=3.11\times 10^-1 m \\ r_2=4.04\times 10^-1 m \\$

Electric force is givenby,

$F= \frac {KQ^2}{r^2}$ K: coulomb`s constant

$F= \frac{1}{r^2}$

Use $F_1r_1^2=F_2r_2^2$

$F_2=\frac {F_1r_1^2}{r^2}$

$F_2= \frac{5.2\times10^-4 N \times (3.11 \times 10^-1)^2 m^2}{(4.04 \times 10^-1)^2m^2}$

Ans:

$F_2=3.08\times 10^{-4} N$