What is the force, in terms of Coulomb's constant, between two electrical charges of #6 C# and #-6 C# that are #12 m # apart?

1 Answer
Jun 3, 2017

Answer:

#F = -1/4k_e#

Explanation:

The electric force #F# between two point charges #q_1# and #q_2# a distance #r# apart is given by the equation

#F =k_e(q_1q_2)/(r^2)#

where #k_e# is Coulomb's constant, equal to #8.988 xx 10^9 ("N • m"^2)/("C"^2)#

Plugging in known values, we have

#F =k_e((6"C")(-6"C"))/((12"m")^2)#

#= k_e(-36"C"^2)/(144"m"^2)#

#= -(1"C"^2)/(4"m"^2)k_e#

So, in terms of Coulomb's constant (#k_e#), the (attractive) electric force acting between the two point charges is

#color(red)(F = -1/4k_e)#

(excluding units)