# A charge of 2 C is at (-3 ,7) and a charge of 3 C is at ( 4 , -7 ). If both coordinates are in meters, what is the force between the charges?

Dec 30, 2015

$F = 2.204 \cdot {10}^{8} N$

#### 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 $\left(- 3 , 7\right)$ and $\left({x}_{2} , {y}_{2}\right)$ represent $\left(4 , - 7\right)$.

$\implies d = \sqrt{{\left(4 - \left(- 3\right)\right)}^{2} + {\left(- 7 - 7\right)}^{2}}$
implies d=sqrt((7)^2+(-14)^2
$\implies d = \sqrt{49 + 196}$
$\implies d = \sqrt{245}$

Hence the distance between the two charges is $\sqrt{245}$.

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 $9 \cdot {10}^{9}$,${q}_{1}$ 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 $r = \sqrt{245}$.

$\implies F = \frac{9 \cdot {10}^{9} \cdot 2 \cdot 3}{\sqrt{245}} ^ 2 = \frac{54 \cdot {10}^{9}}{245} = 0.2204 \cdot {10}^{9} N = 2.204 \cdot {10}^{8} N$
$\implies F = 2.204 \cdot {10}^{8} N$
This force is repulsive.