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

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Answer 1 · 2017-01-27T21:17:40

According to Coulumbs Law: $mathbf F = (k Q_1Q_2)/(abs(\mathbf r_1 - \mathbf r_2)^2) mathbf e_(r_1 - r_2)$

Coulumb's Constant is $k approx 8.99×10^9 Nm^2$ / $C^2$

The magnitude of the force evaluates as:

$| mathbf F |= ( 8.99×10^9 * (-2) * (-1))/(abs(((5),(3)) - ((3),(1)))^2) = 2.2475 * 10^9 N$

....with $abs(\mathbf r_1 - \mathbf r_2)^2 = 8$ .

[The answer may look big but note that a charge of "only" 1C is a lot of charge.]

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