What is the force, in terms of Coulomb's constant, between two electrical charges of #-2 C# and #-3 C# that are #6 m # apart?

Question

Elena Albarran · Accepted Answer

#F=1,5*10^9 N# #F=(k*q_1*q_2)/r^2# #F=(9*10^9*(-2)*(-3))/6^2# #F=(9*6*10^9)/36# #F=1,5*10^9 N#

Adrian Ayala · Answer

undefined The force between two electrical charges can be calculated using Coulomb's law, which states that the force $ F $ between two point charges $ q_1 $ and $ q_2 $ separated by a distance $ r $ is given by:

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

Where $ k $ is Coulomb's constant, with a value of approximately $ 8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 $.

Given $ q_1 = -2 \, \text{C} $, $ q_2 = -3 \, \text{C} $, and $ r = 6 \, \text{m} $, we can calculate the force:

$$ F = \frac{{8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \cdot |-2 \, \text{C}| \cdot |-3 \, \text{C}|}}{{(6 \, \text{m})^2}} $$

$$ F = \frac{{8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \cdot 2 \, \text{C} \cdot 3 \, \text{C}}}{{36 \, \text{m}^2}} $$

$$ F = \frac{{8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \cdot 6 \, \text{C}^2}}{{36 \, \text{m}^2}} $$

$$ F = \frac{{53.925 \times 10^9 \, \text{N} \cdot \text{m}^2}}{{36 \, \text{m}^2}} $$

$$ F \approx \frac{{53.925 \times 10^9}}{{36}} \, \text{N} $$

$$ F \approx 1.497 \times 10^9 \, \text{N} $$

Therefore, the force between the two charges is approximately $ 1.497 \times 10^9 \, \text{N} $.