Question

Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m. A positive point charge q3 is located at the origin. a) What must the value of q3 be for the net force on thispoint charge to have a magnitude 4.40 μN ?

Answer 1

q₃=5.3nC

Step-by-step explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

`F= k(|q_aq_b|)/(r^(2) ) \\\\\\F_(13)=(9*10^(9) Nm^(2) /C^(2) )(|(-4.00*10^(-9)C)q_3|)/((.250m)^(2) ) =576q_3N/C\\\\F_(23)=(9*10^(9) Nm^(2) /C^(2) )(|(2.40*10^(-9)C)q_3|)/((.300m)^(2) ) =240q_3N/C\\`

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

`F_(Net)=F_(13)+F_(23)\\\\4.40*10^(-9) N=576q_3N/C+240q_3N/C\\\\4.40*10^(-6) N=816q_3N/C\\\\\implies q_3=5.3*10^(-9)C=5.3nC`

In words, the value of q₃ must be 5.3nC.