Answer suppose to be something similar to ex000246N In that

Answer suppose to be something similar to ex:0.00246N! In that type of form.

What is the force on the charge located at x = +8.00 cm in Figure 17.40(a) given that q = 5.00 nuC and a = 7.50? (The positive direction is to the right.)

Solution

Coulombs law: F = k *Q1*Q2/r² , F (N), Q (C), k= 9*10^9 , r (m)
Q1, Q2 different sign => attraction force

Number the charges from left to right 1, 2, 3.

1 ->2 F1 = k*q*(-aq)/r1² , attraction
3->2 F3 = k*q*(-aq)/r3² , attraction =>

Net force on charge 2: F3 - F1 , to the right because shorter distance=> stronger force.
F3 = 9*10^9 * (5.0*10)²/ (7*10²)²
F3 =45.91 N

F1 = 9*10^9 * (5.0*10)²/ (5*10²)² = 0.992

Fnet = 45.91- 4.5 = 4.1*10^-3 N