Question

what is the electric force between a glass ball that has 2.5 uc charge and a rubber ball that has a -1.5 uc charge when they are separated by a distance of 5.0 cm

Answer 1

F= k/q₁/q₂/d₂
F = 9 × 10⁹ ( 2.5 × 10⁻⁶)/0.05²

= 45N

Answer 2

Force between two charges is given by

$F = \frac{kq_1q_2}{r^2}$

here we know that

$k = 9*10^9$

$q_1 = 2.5\mu C$

$q_2 = -1.5 \mu C$

$r = 5 cm = 0.05 m$

now using the above equation we can say

$F = \frac{9*10^9* 2.5*10^{-6} * (-1.5*10^{-6})}{0.05^2}$

$F = -13.5 N$

so here force will be attraction force between two balls and the magnitude of force will be 13.5 N