# What is the force, in terms of Coulomb's constant, between two electrical charges of -45 C and 30 C that are 15 m  apart?

Jan 12, 2016

-6k or $- 5.4 \times {10}^{10} N$

#### Explanation:

You would use Coulomb's Law for this question:

${F}_{e} = \frac{k {q}_{1} {q}_{2}}{{r}^{2}}$
where q1 represents the quantity of charge on object 1 (in Coulombs), q2 represents the quantity of charge on object 2 (in Coulombs), and r represents the distance of separation between the two objects (in meters). "k" is a constant, and it's value is $9.0 \times {10}^{9} \frac{N {m}^{2}}{C} ^ 2$

If we plug in the values we get:

${F}_{e} = \frac{k \left(- 45 C\right) \left(30 C\right)}{{\left(15 m\right)}^{2}}$

In terms of "k", this would simplify to:

${F}_{e} = k \left(- 6\right)$

If we sub in the value for k, we get:

${F}_{e} = \left(9.0 \times {10}^{9} \frac{N {m}^{2}}{C} ^ 2\right) \left(- 6\right)$

${F}_{e} = - 5.4 \times {10}^{10} N$

A negative force indicates a force of ATTRACTION, whereas a positive force indicates a force of REPULSION.

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