Question

Find the electric force between charges of -2.0 x 10 C and +5.0 x 10 C which are separated by a distance of 0.75m in a vacuum.

Answer 1

By Coulomb's Law

$$|F_e|=k_e\dfrac{|q_1||q_2|}{\varepsilon r^2}$$

Given $k_e=8.988×10^{9} N⋅m^2⋅C^{−2},$

$q_1=-2.0\times10^{-6}C, q_2=+5.0\times10^{-6}C,$

$r=0.75\ m, \varepsilon=1$

$$|F|=8.988×10^{9} N⋅m^2⋅C^{−2}\cdot\dfrac{|-2.0\times10^{-6}C||+5.0\times10^{-6}C|}{1(0.75\ m) ^2}$$

$$=0.16\ N, attractive$$