Question

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?

Answer 1

Each charge experiences a repulsive force of `F= 1.01xx10^8N`

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|/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 = (3m)^2+(13m)^2 = 178m^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.

Advanced:

The constant `k_e` can be expressed in terms of the SI constant, the permittivity of free space `epsilon_o`, as:

`k_e=1/(4 pi epsilon_o)`