# A charge of -2 C is at (4, 7) and a charge of -1 C is at ( 5, 2). If both coordinates are in meters, what is the force between the charges?

Jun 24, 2018

Charge ${q}_{1} = - 2 C$ is placed at $\left(4 , 7\right)$
Charge ${q}_{2} = - 1 C$ is placed at $\left(5 , 2\right)$

Distance between both charges is
r=sqrt((5-4)^2+(2-7)^2
r=sqrt((1)^2+(-5)^2
r=sqrt(1+25
$r = \sqrt{26} m$ and ${r}^{2} = 26$

Force between two charges ${q}_{1}$ and ${q}_{2}$ is
$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$ where $k = 9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$
=>F=((9*10^9)(-2)(-1))/(26
=>F=((9*10^9)(2))/(26
$\implies F = \frac{\left(9 \cdot {10}^{9}\right)}{13} N$
$\implies F = 0.6923 \cdot {10}^{9} N$
or
$\implies F = 6.923 \cdot {10}^{8} N$