Respuesta :
Force between two charges is given by
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know that
[tex]k = 9*10^9[/tex]
[tex]q_1 = 2.5\mu C[/tex]
[tex]q_2 = -1.5 \mu C[/tex]
[tex]r = 5 cm = 0.05 m[/tex]
now using the above equation we can say
[tex]F = \frac{9*10^9* 2.5*10^{-6} * (-1.5*10^{-6})}{0.05^2}[/tex]
[tex]F = -13.5 N[/tex]
so here force will be attraction force between two balls and the magnitude of force will be 13.5 N
Force on the particle is defined as the application of the force field of one particle on another particle. The electric force between a glass ball will be -13.5 N.
What is electric force?
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The given data in the problem is;
F is the electric force between a glass ball=?
Q₁ is the megnitude of charge 1=2.5 μC=2.5×10⁻⁶C
Q₂ is the megnitude of charge 2=-1.5 uc=1.5×10⁻⁶C
r is the separated distance =5.0 cm
The formula for the electric force is given by;
[tex]\rm F=K\frac{KQ_1Q_2}{r^2} \\\\ \rm F=\frac{9\times10^9\times2.5\times10^{-6}-1.5\times10^{-6}}{(5\times10^{-2})^2}\\\\ \rm F=-13.5 N[/tex]
Hence the electric force between a glass ball will be -13.5 N.
To learn more about the electric force refer to the link;
https://brainly.com/question/1076352
