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

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

Step-by-step 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) *(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 (+)