A charge of $-2 C$ is at $(6 , -5 )$ and a charge of $-3 C$ is at $(3, 5 )$ . If both coordinates are in meters, what is the force between the charges?

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A charge of $-2 C$ is at $(6 , -5 )$ and a charge of $-3 C$ is at $(3, 5 )$ . If both coordinates are in meters, what is the force between the charges?

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Answer 1 · 2015-12-30T11:58:28

$F=4.95*10^8N$

Explanation:

The distance formula for Cartesian coordinates is

$d=sqrt((x_2-x_1)^2+(y_2-y_1)^2$
Where $x_1, y_1$ , and $x_2, y_2$ are the Cartesian coordinates of two points respectively.
Let $(x_1,y_1)$ represent $(6,-5)$ and $(x_2,y_2)$ represent $(3,5)$ .

$implies d=sqrt((3-6)^2+(5-(-5))^2)$
$implies d=sqrt((-3)^2+(10)^2$
$implies d=sqrt(9+100)$
$implies d=sqrt(109)$

Hence the distance between the two charges is $sqrt(109)$ .

The electrostatic force between two charges is given by
$F=(kq_1q_2)/r^2$
Where $F$ is the force between the charges, $k$ is the constant

and its value is $9*10^9 Nm^2/C^2$ , $q_1$ and $q_2$ are the

magnitudes of the charges and $r$ is the distance between the two charges.

Here $F=??$ , $k=9*10^9Nm^2/C^2$ , $q_1=-2C$ , $q_2=-3C$ and $r=d=sqrt109$ .

$implies F=(9*10^9(-2)(-3))/(sqrt109)^2=(9*10^9*6)/109=(54*10^9)/109=0.495*10^9N=4.95*10^8N$
$implies F=4.95*10^8N$

Since the force is positive therefore the force is repullsive.

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A charge of $-2 C$ is at $(6 , -5 )$ and a charge of $-3 C$ is at $(3, 5 )$ . If both coordinates are in meters, what is the force between the charges?

1 Answer

Explanation:

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A charge of $-2 C$ is at $(6 , -5 )$ and a charge of $-3 C$ is at $(3, 5 )$ . If both coordinates are in meters, what is the force between the charges?