Question

A 7.94-nC charge is located 1.77 m from a 4.14-nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. (b) is the force attractive or repulsive?

Answer 1

Answer:

F=94.32*10⁻⁹N  , The force F is repusilve because both charges have the same sign (+)

Explanation:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters(m)

Equivalence  

1nC= 10⁻⁹C

Data

K=8.99x10⁹N*m²/C²

q₁ = 7.94-nC= 7.94*10⁻⁹C

q₂= 4.14-nC=  4.14 *10⁻⁹C

d= 1.77 m

Magnitude of the electrostatic force that one charge exerts on the other

We apply formula (1):

$F=8.99x10^{9} *\frac{7.94*10^{-9} *4.14 *10^{-9} }{1.77^{2} }$

F=94.32*10⁻⁹N  , The force F is repusilve because both charges have the same sign (+)