A charge of $- 2 C$ $-2 C$ is at $(- 5, 1)$ $(-5 , 1 )$ and a charge of $- 3 C$ $-3 C$ is at $(0, 4)$ $(0 ,4)$ . If both coordinates are in meters, what is the force between the charges?

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A charge of $- 2 C$ $-2 C$ is at $(- 5, 1)$ $(-5 , 1 )$ and a charge of $- 3 C$ $-3 C$ is at $(0, 4)$ $(0 ,4)$ . If both coordinates are in meters, what is the force between the charges?

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Answer 1 · 2017-05-20T01:53:58

Magnitude of $\to F$ $vecF$ : $4.46 \times 10^{10} N$ $4.46 xx 10^10 N$

Direction of ${\to F}_{1 on 2}$ $vecF_(1"on"2)$ : ${31.0}^{o}$ $31.0^o$

Direction of ${\to F}_{2 on 1}$ $vecF_(2"on"1)$ : ${211.0}^{o}$ $211.0^o$

Explanation:

We can solve this equation by using Coulomb's law, which relates the magnitude of the electric force between two point charges, and in equation form is

$F = \frac{1}{4 π ε_{0}} \frac{| q_{1} q_{2} |}{r^{2}}$ $F = 1/(4piepsilon_0)(|q_1q_2|)/(r^2)$

The quantity $\frac{1}{4 π ε_{0}}$ $1/(4piepsilon_0)$ is a proportionality constant, and can sometimes be simply called $k$ $k$ , and has a value of $\frac{1}{4 π (8.854 \times 10^{- 12} \frac{C^{2}}{N \cdot m^{2}})}$ $1/(4pi(8.854 xx 10^-12(C^2)/(N*m^2)))$

$= 8.988 \times 10^{9} \frac{N \cdot m^{2}}{C^{2}}$ $= 8.988 xx 10^9(N*m^2)/(C^2)$

The variables $q_{1}$ $q_1$ and $q_{2}$ $q_2$ are, in no particular order, the magnitude of the electric charges of the two point charges, which are $- 2 C$ $-2C$ and $- 3 C$ $-3C$ . The variable $r$ $r$ is the distance between the two point charges, which is

$∣ ∣ ∣ \sqrt{{(- 5 m)}^{2} + {(1 m)}^{2}} - \sqrt{{(0 m)}^{2} + {(4 m)}^{2}} ∣ ∣ ∣ = 1.10 m$ $|sqrt((-5m)^2 + (1m)^2) - sqrt((0m)^2 + (4m)^2)| = 1.10m$

Now that we have all the variables needed, the magnitude of the electric force between the two points is

$F = 8.988 \times 10^{9} \frac{N \cdot m^{2}}{C^{2}} (\frac{| (- 2 C) (- 3 C) |}{{(1.10 m)}^{2}})$ $F = 8.988 xx 10^9(N*m^2)/(C^2)((|(-2C)(-3C)|)/((1.10m)^2))$

$= 4.46 \times 10^{10} N$ $= color(blue)(4.46 xx 10^10 N$

Since both point charges are negative, the force is repulsive, and is the magnitude of the force the particles act upon each other.

If you want the direction of the force, which completes the vectoral description of the force of repulsion, we can use the fact that the force of object $1$ $1$ acting on $2$ $2$ is equal to the negative of the force of object $2$ $2$ acting on $1$ $1$ .

If we let the point with charge $- 2 C$ $-2C$ with coordinates $(- 5, 1)$ $(-5,1)$ be $1$ $color(orange)1$ , and the point with charge $- 3 C$ $-3C$ with coordinates $(0, 4)$ $(0,4)$ be $2$ $color(red)2$ (for convenience), the direction of ${\to F}_{1 on 2}$ $vec F_(color(orange)1"on"color(red)2)$ is

$arctan (\frac{y_{2} - y_{1}}{x_{2} - x_{1}}) = arctan (\frac{4 m - 1 m}{0 m - - 5 m}) = {31.0}^{o}$ $arctan((y_2-y_1)/(x_2-x_1)) = arctan((4m-1m)/(0m--5m)) = color(green)(31.0 ^o$

The direction of ${\to F}_{2 on 1}$ $vec F_(2"on"1)$ is simply the direction opposite to this in a circle:

${180.0}^{o} + {31.0}^{o} = {211.0}^{o}$ $180.0^o + color(green)(31.0^o) = color(purple)(211.0^o$

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A charge of $- 2 C$ $-2 C$ is at $(- 5, 1)$ $(-5 , 1 )$ and a charge of $- 3 C$ $-3 C$ is at $(0, 4)$ $(0 ,4)$ . If both coordinates are in meters, what is the force between the charges?

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Explanation:

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A charge of −2C-2 C is at (−5,1)(-5 , 1 ) and a charge of −3C-3 C is at (0,4)(0 ,4) . If both coordinates are in meters, what is the force between the charges?

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Explanation:

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A charge of $- 2 C$ $-2 C$ is at $(- 5, 1)$ $(-5 , 1 )$ and a charge of $- 3 C$ $-3 C$ is at $(0, 4)$ $(0 ,4)$ . If both coordinates are in meters, what is the force between the charges?