Three point charges are arranged on a line. charge q3 = +5.00 nc and is at the origin. charge q2 = -2.00 nc and is at x = 5.00 cm . charge q1 is at x = 2.50 cm .what is q1 (magnitude and sign) if the net force on q3 is zero?

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

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

o solve this problem we apply Coulomb's law:  

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

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm=  2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

We eliminate k*q₃ of the equation

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC