A charge of #5 C# is at #(-6, 1 )# and a charge of #-3 C# is at #(-2, 1) #. If both coordinates are in meters, what is the force between the charges?

1 Answer
Mar 20, 2016

Answer:

The force between the charges is #8\times10^9# N.

Explanation:

Use Coulomb's law:
#F=\frac{k\abs{q_1q_2}}{r^2}#

Calculate #r#, the distance between the charges, using the Pythagorean theorem
#r^2 = \Delta x^2 + \Delta y^2#
#r^2 = (-6-(-2))^2 + (1-1)^2#
#r^2 = (-6+2)^2 + (1-1)^2#
#r^2 = 4^2 + 0^2#
#r^2 = 16#
#r=4#

The distance between the charges is #4#m. Substitute this into Coulomb's law. Substitute in the charge strengths as well.

#F=\frac{k\abs{q_1q_2}}{r^2}#
#F=k\frac{\abs{(5)(-3)}}{4^2}#
#F=k\frac{15}{16}#
#F = 8.99×10^9(\frac{15}{16})# (Substitute in the value of Coulomb's constant)
#F=8.4281\times 10^9# N
#F=8 \times 10^9# N (As you're working with one significant figure)