# A charge of 1 C is at (-2,7,1) and a charge of 2 C is at (0,-4 ,4) . If both coordinates are in meters, what is the force between the charges?

$1.3433 \setminus \times {10}^{8} \setminus N$

#### Explanation:

The distance $d$ between two points $\left(- 2 , 7 , 1\right)$ & $\left(0 , - 4 , 4\right)$ is given by distance formula

$d = \setminus \sqrt{{\left(- 2 - 0\right)}^{2} + {\left(7 - \left(- 4\right)\right)}^{2} + {\left(1 - 4\right)}^{2}} = \setminus \sqrt{134}$

Now, the electrostatic force $F$ between the point charges ${Q}_{1} = 1 C$ & ${Q}_{2} = 2 C$ at a distance $d = \setminus \sqrt{134}$

$F = \setminus \frac{1}{4 \setminus \pi \setminus {\epsilon}_{0}} \setminus \frac{{Q}_{1} {Q}_{2}}{{d}^{2}}$

$= \left(9 \setminus \times {10}^{9}\right) \setminus \frac{1 \setminus \cdot 2}{{\left(\setminus \sqrt{134}\right)}^{2}}$

$= 1.3433 \setminus \times {10}^{8} \setminus N$