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

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

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Answer 1 · 2016-01-12T14:34:00

-6k or $- 5.4 \times 10^{10} N$ $-5.4xx10^10N$

Explanation:

You would use Coulomb's Law for this question:

$F_{e} = \frac{k q_{1} q_{2}}{r^{2}}$ $F_e=(kq_1q_2)/(r^2)$
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 $9.0 \times 10^{9} \frac{N m^{2}}{C^{2}}$ $9.0xx10^(9)(Nm^2)/C^2$

If we plug in the values we get:

$F_{e} = \frac{k (- 45 C) (30 C)}{{(15 m)}^{2}}$ $F_e=(k(-45C)(30C))/((15m)^2)$

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

$F_{e} = k (- 6)$ $F_e=k(-6)$

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

$F_{e} = (9.0 \times 10^{9} \frac{N m^{2}}{C^{2}}) (- 6)$ $F_e=(9.0xx10^(9)(Nm^2)/C^2)(-6)$

$F_{e} = - 5.4 \times 10^{10} N$ $F_e=-5.4xx10^10N$

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

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

1 Answer

Explanation:

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What is the force, in terms of Coulomb's constant, between two electrical charges of −45C-45 C and 30C30 C that are 15m15 m apart?

1 Answer

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