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

1440 N

Step-by-step explanation:

The electrostatic force between the two charges is given by:

`F=k(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) )((-8.00 \cdot 10^(-5)C)^2)/((0.20 m)^2)=1440 N`