# A charge of -2 C is at (4, 7) and a charge of -1 C is at ( 1 , -6 ). If both coordinates are in meters, what is the force between the charges?

Mar 2, 2016

Each charge experiences a repulsive force of $F = 1.01 \times {10}^{8} N$

#### Explanation:

Let's start by creating a diagram of the information that we know:

We know from Coulomb's Law that the force between charges depends on the size and sign of the charges, and on the distance between them:

$F = {k}_{e} | {q}_{1} {q}_{2} \frac{|}{r} ^ 2$

Where ${q}_{1}$ and ${q}_{2}$ are the two charges, ${k}_{e}$ is Coulomb's constant (k_e = 8.99xx10^9 N m^2 C^(−2)), and $r$ is the distance between the charges. We see that we need to calculate ${r}^{2}$ which can be gotten by using Pythagoras's Theorem on the right triangle shown in the diagram:

${r}^{2} = \Delta {x}^{2} + \Delta {y}^{2} = {\left(3 m\right)}^{2} + {\left(13 m\right)}^{2} = 178 {m}^{2}$

We can now get the force:

F= 8.99xx10^9 N m^2 C^(−2) * |(-2C)*(-1C)|/(178m^2)=1.01xx10^8N

Each charge experiences this force which is repulsive, since they have the same sign of charge.

The constant ${k}_{e}$ can be expressed in terms of the SI constant, the permittivity of free space ${\epsilon}_{o}$, as:
${k}_{e} = \frac{1}{4 \pi {\epsilon}_{o}}$