# 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?

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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} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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