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?

1 Answer
Sep 13, 2017

Answer:

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

Explanation:

Coulomb's Law, in equation form, 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.

The force is a force of attraction if F has a negative value and repulsion if the value of F has a positive value.

#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)#.

My apologies for the need to edit, I originally overlooked the request that the answer be in terms of k."

I hope this helps,
Steve