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Bogdan [553]
4 years ago
15

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.
Physics
1 answer:
ss7ja [257]4 years ago
5 0

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

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