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Two identical charges, each -8.00 E-5 C, are separated by a distance of 20.0 cm (100 cm = 1 m). What is the force of repulsion? Coulomb's constant is 9.00 E9 N*m2/C2


Remember to identify all data (givens and unknowns), list equations used, show all your work, and include units and the proper number of significant digits to receive full credit.

Respuesta :

Answer:

1440 N

Explanation:

The electrostatic force between the two charges is given by:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where:

[tex]k=9.00 \cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb's constant

[tex]q_1 = q_2 = -8.00 \cdot 10^{-5} C[/tex] is the value of each charge

r = 20.0 cm = 0.20 m is the separation between the charges

Substituting numbers into the equation, we find

[tex]F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(-8.00 \cdot 10^{-5}C)^2}{(0.20 m)^2}=1440 N[/tex]