# 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?

Dec 30, 2015

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

#### 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 $\left({x}_{1} , {y}_{1}\right)$ represent $\left(6 , - 5\right)$ and $\left({x}_{2} , {y}_{2}\right)$ represent $\left(3 , 5\right)$.

$\implies d = \sqrt{{\left(3 - 6\right)}^{2} + {\left(5 - \left(- 5\right)\right)}^{2}}$
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
$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$
Where $F$ is the force between the charges, $k$ is the constant

and its value is $9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$,${q}_{1}$ and ${q}_{2}$ are the

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

Here F=??, $k = 9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$, ${q}_{1} = - 2 C$, ${q}_{2} = - 3 C$ and $r = d = \sqrt{109}$.

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

Since the force is positive therefore the force is repullsive.