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

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Answer 1 · 2015-12-30T11:58:28

$F = 4.95 \cdot 10^{8} N$ $F=4.95*10^8N$

Explanation:

The distance formula for Cartesian coordinates is

$d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $d=sqrt((x_2-x_1)^2+(y_2-y_1)^2$
Where $x_{1}, y_{1}$ $x_1, y_1$ , and $x_{2}, y_{2}$ $x_2, y_2$ are the Cartesian coordinates of two points respectively.
Let $(x_{1}, y_{1})$ $(x_1,y_1)$ represent $(6, - 5)$ $(6,-5)$ and $(x_{2}, y_{2})$ $(x_2,y_2)$ represent $(3, 5)$ $(3,5)$ .

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

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

The electrostatic force between two charges is given by
$F = \frac{k q_{1} q_{2}}{r^{2}}$ $F=(kq_1q_2)/r^2$
Where $F$ $F$ is the force between the charges, $k$ $k$ is the constant

and its value is $9 \cdot 10^{9} N \frac{m^{2}}{C^{2}}$ $9*10^9 Nm^2/C^2$ , $q_{1}$ $q_1$ and $q_{2}$ $q_2$ are the

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

Here $F = ? ?$ $F=??$ , $k = 9 \cdot 10^{9} N \frac{m^{2}}{C^{2}}$ $k=9*10^9Nm^2/C^2$ , $q_{1} = - 2 C$ $q_1=-2C$ , $q_{2} = - 3 C$ $q_2=-3C$ and $r = d = \sqrt{109}$ $r=d=sqrt109$ .

$\Rightarrow F = \frac{9 \cdot 10^{9} (- 2) (- 3)}{{(\sqrt{109})}^{2}} = \frac{9 \cdot 10^{9} \cdot 6}{109} = \frac{54 \cdot 10^{9}}{109} = 0.495 \cdot 10^{9} N = 4.95 \cdot 10^{8} N$ $implies F=(9*10^9(-2)(-3))/(sqrt109)^2=(9*10^9*6)/109=(54*10^9)/109=0.495*10^9N=4.95*10^8N$
$\Rightarrow F = 4.95 \cdot 10^{8} N$ $implies F=4.95*10^8N$

Since the force is positive therefore the force is repullsive.

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

1 Answer

Explanation:

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