# Two charges of  1 C  and  -2 C  are at points  (0,4,2)  and  ( 5,3,7), respectively. Assuming that both coordinates are in meters, what is the force between the two points?

Apr 17, 2017

$F = - 3.5 \cdot {10}^{8} \text{ Newtons}$

#### Explanation:

$\text{First,you should find the distance between two point given.}$

${r}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}$

${P}_{1} \left({x}_{1} , {y}_{1} , {z}_{1}\right) = \left(0 , 4 , 2\right)$
${P}_{2} \left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(5 , 3 , 7\right)$

${r}^{2} = {\left(5 - 0\right)}^{2} + {\left(3 - 4\right)}^{2} + {\left(7 - 2\right)}^{2}$
${r}^{2} = 25 + 1 + 25$
${r}^{2} = 51$

$\text{Now ,we can use the Coulomb's law to calculate force between two charge.}$

$F = K \cdot \frac{{Q}_{1} \cdot {Q}_{2}}{r} ^ 2$

$\text{Where "Q_1=1C" , "Q_2=-2C" , "r^2=51" , } K = 9 \cdot {19}^{9}$

$F = \frac{{9.10}^{9} \cdot 1 \cdot \left(- 2\right)}{51}$

$F = - 0.35 \cdot {10}^{9} \text{ Newtons}$

$\text{The minus sign means that there is an attractive force between two charges.}$