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

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)