Question

Consider two point charges located on the x axis: one charge, q1 = -15.0 nC , is located at x1 = -1.725 m ;the second charge, q2 = 38.0 nC , is at the origin (x=0.0000). What is the net force exerted by these two charges on a third charge q3 = 46.0 nC placed between q1 and q2 at x3 = -1.075 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer 1

Answer:

Fn₃= -28.3*10⁻⁶N  (in direction -x)  :net force exerted by  two charges q₁,q₂  on a third charge q₃.

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces .

Equivalences

1nC= 10⁻⁹ C

Known data

k= 9 *10⁹ N*m² /C²

q₁=-15.0 nC=-15*10⁻⁹ C

q₂=+38 nC =+38*10⁻⁹ C  

q₃= +46 nC= =+46*10⁻⁹ C  

d₁₃= 0.65 m

d₂₃= 1.075 m

d₁₃: distance from q₁  to q3

d₂₃: distance from q₂ to q3

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force F₂₃ of q₂ on q₃ is repulsive because the charges have equal signs and the forces : Force F₂₃ is directed to the left (-x).

The force F₁₃ of q₁ on q₃ is attractive because the charges have opposite signs. Force F₁₃ is directed to the left (-x)

Calculation of the net force exerted for q₁ and q₂ on the charge q₃

The net force exerted for q₁ and q₂ on the charge q₃ is the algebraic sum of the forces F₁₁₃ and F₂₃ because both acts on the x-axis.

Fn₃= F₁₃+F₂₃

To calculate the magnitudes of the forces exerted by the charges q₁, and q₂ on q₃ we apply Coulomb's law:

F₁₃=(k*q₁*q₃)/d₁₃²  

F₁₃=(9*10⁹*15*10⁻⁹*46*10⁻⁹)/(0.65)² = 14698.2*10⁻⁹ N = 14.7*10⁻⁶N

F₁₃= 14.7*10⁻⁶N, in the direction of the x-axis negative (-x)

F₃=(k*q₂*q₃)/d₂₃²

F₂₃=(9*10⁹*38*10⁻⁹*46*10⁻⁹)/(1.075)² =13623.4*10⁻⁹ N= 13.6*10⁻⁶N

F₂₃=13.6*10⁻⁶N (in direction -x)

Net force on q₃

Fn₃= -14.7*10⁻⁶N - 13.6*10⁻⁶N = -28.3*10⁻⁶N  

Fn₃= -28.3*10⁻⁶N  (in direction -x)