The force of attraction between two point masses is 10-4N when the distance between them is 0.18m. If the distance is reduced to 0.06m, calculate the force.
-
A.
1.1 x 10-5N -
B.
3.3 x 10-5N -
C.
3.0 x 10-4N -
D.
9.0 x 10-4N
Correct Answer: Option D
Explanation
F (alpha frac{1}{r^2})
therefore, (frac{F_1}{F_2}) = (frac{r_1}{(r_2)^2})
(frac{10^{-4}}{F_2}) = (frac{(0.06)^2}{(0.18)^2})
F = 9 x 10-4N