What is the force, in terms of Coulomb's constant, between two electrical charges of #22 C# and #-32 C# that are #6 m # apart?

Question

Adrian Ayala · Accepted Answer

It is an attraction force of #1.76*10^11 N#.

Coulomb's Law expressed as an equation is #F = k*(q_1*q_2)/r^2# where F is the magnitude of the force of attraction, or repulsion, between 2 charges, k is known as Coulomb's constant and has a value of #8.99×10^9 (N m^2)/C^2#, #q_1 and q_2# are the 2 charges, in Coulombs, and r is the distance between the charges, in meters. If the value of F is negative, the force is one of attraction; otherwise, it is one of repulsion. #color(green)(Edit " The green lines between here and the red text below replace those red lines.")# #color(green)("Plugging the values into the formula")# #color(green)(F = k * (22 C*(-32 C))/(6 m)^2)# #color(green)(F = -k * 19.6 N)# #color(green)("So it is an attraction force of " k * 19.6 N)#. #color(red)("Plugging the values into the formula")# #color(red)(F = (8.99×10^9 (N m^2)/C^2) * (22 C*(-32 C))/(6 m)^2)# #color(red)(F = -175.8*10^9 N = -1.76*10^11 N)# #color(red)("So it is an attraction force of " 1.76*10^11 N)#. I apologize for the editing that was necessary; I had not noticed at first that the response was to be in terms of k." Hope this is helpful, Steve.

Amelia Adams · Answer

undefined The force between two electrical charges can be calculated using Coulomb's law:

$F = \frac{{k \cdot |q_1 \cdot q_2|}}{{r^2}}$

Where:
- $F$ is the force between the charges,
- $k$ is Coulomb's constant ($8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2$),
- $q_1$ and $q_2$ are the magnitudes of the charges, and
- $r$ is the distance between the charges.

Plugging in the given values:

$F = \frac{{(8.9875 \times 10^9) \cdot |22 \, \text{C} \cdot (-32 \, \text{C})|}}{{(6 \, \text{m})^2}}$