# A charge of 1 C is at (3,-2) and a charge of -3 C is at ( 7,-1) . If both coordinates are in meters, what is the force between the charges?

Mar 13, 2016

$F = - 1 , 59 \cdot {10}^{9} \text{ } N$

#### Explanation:

$F = \text{force between two charges}$
${q}_{1} , {q}_{2} : \text{charges}$
$d : \text{distance between two charges}$
$\text{The force between two charges is determined by :}$
$F = k \frac{{q}_{1} \cdot {q}_{2}}{d} ^ 2$
$\text{let's calculate distance between two charges}$
${P}_{1} = \left(3 , - 2\right)$
${P}_{2} = \left(7 , - 1\right)$
$\Delta x = {P}_{\text{2x"-P_"1x}} = 7 - 3 = 4$
$\Delta y = {P}_{\text{2y"-P_"1y}} = - 1 + 2 = 1$
$d = \sqrt{\Delta {x}^{2} + \Delta {y}^{2}} \text{ "d=sqrt(4^2+1^2)" } d = \sqrt{17}$
$F = k \frac{1 \cdot \left(- 3\right)}{\sqrt{17}} ^ 2$
$F = \frac{- 3 k}{17} \text{ } k = 9 \cdot {10}^{9} N \cdot {m}^{2} \cdot {C}^{-} 2$
$F = \frac{- 3 \cdot 9 \cdot {10}^{9}}{17}$
$F = - \frac{27 \cdot {10}^{9}}{17}$
$F = - 1 , 59 \cdot {10}^{9} \text{ } N$