# A charge of 2 C is at (5, 2) and a charge of -3 C is at ( 7,-1) . If both coordinates are in meters, what is the force between the charges?

Apr 10, 2016

$F = - 4 , 15 \cdot {10}^{9} \text{ N}$

#### Explanation:

$\text{Force between two charges is given by Coulomb's law }$

$\text{given data:}$
${q}_{1} = 2 C$
${q}_{2} = - 3 C$
${P}_{1} = \left(5 , 2\right)$
P_2=(7,-1))
$k = 9 \cdot {10}^{9} N \cdot {m}^{2} \cdot {C}^{-} 2$
F=?

$\text{formula :}$

$F = k \cdot \frac{{q}_{1} \cdot {q}_{2}}{d} ^ 2$

$\text{distance between two points is given by :}$
$d = \sqrt{{\left(7 - 5\right)}^{2} + {\left(- 1 - 2\right)}^{2}}$

$d = \sqrt{{2}^{2} + {\left(- 3\right)}^{2}}$

$d = \sqrt{4 + 9} = \sqrt{13}$
${d}^{2} = 13 \text{ m}$

$F = 9 \cdot {10}^{9} \frac{2 \cdot \left(- 3\right)}{13}$

$F = - \frac{54}{13} \cdot {10}^{9}$

$F = - 4 , 15 \cdot {10}^{9} \text{ N}$