The magnitude of the electrostatic force between the two charges is [tex]1.12\cdot 10^{14} N[/tex]
Explanation:
The magnitude of the electrostatic force between two charges is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
In this problem, we have:
[tex]q_1 = +10.0 C[/tex]
[tex]q_2 = -50.0 C[/tex]
r = 20.0 cm = 0.20 m
Therefore, the force between the charges is
[tex]F=(8.99\cdot 10^9) \frac{(10)(-50)}{(0.20)^2}=-1.12\cdot 10^{14} N[/tex]
where the negative sign means that the force is attractive, since the two charges have opposite sign.
Learn more about electrostatic force here:
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