Question

Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of q2 on q1? (Express your answer in vector form.)

Answer 1

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Step-by-step explanation:

F₂₁ =
`(K|q_1|q_2|)/(r^2).(r_2_1)/(|r_2_1|)`

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| =
`√((4^2)+(11.5^2)) = √(148.25)`

(|r₂₁|)² = 148.25

`F_2_1=(K|q_1|q_2|)/(r^2).(r_2_1)/(|r_2_1|) = (8.99X10^9(14X10^(-6))(60X10^(-6)))/(148.25).((4i + 11.5j))/(√(148.25) )`

= 0.050938(0.19107i + 0.54933j) N

= (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N