# A charge of 1 C is at (-5 ,-2 ,7 ) and a charge of 2 C is at (0,8,4) . If both coordinates are in meters, what is the force between the charges?

Feb 13, 2016

Force between charges is 1.343⋅10^8 Newton

#### Explanation:

Distance $d$ between $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is given by

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

In the instant case, this turns out to be $\sqrt{{\left(0 + 5\right)}^{2} + {\left(8 + 2\right)}^{2} + {\left(4 - 7\right)}^{2}}$, which simplifies to $\sqrt{25 + 100 + 9}$ or $\sqrt{134}$.

The force between two charges ${C}_{1}$ and ${C}_{2}$ is given by $k \cdot \frac{{C}_{1} \cdot {C}_{2}}{d} ^ 2$, where $k$ is Coulomb's law constant for the medium between the charges and for air it is $9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$, where charge is in Coulomb and distance in meters.

Assuming it to be so Force will be given by

$9 \cdot 1 \cdot \frac{2}{134} \cdot {10}^{9}$ or $1.343 \cdot {10}^{8}$ Newton.