Question

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.

Answer 1

Answer:

1440 N

Explanation:

The electrostatic force between the two charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where:

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

$q_1 = q_2 = -8.00 \cdot 10^{-5} C$ 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

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