Question: Three point charges are arranged along the x axis. Charge q 1 = -4.40

Question

Question: Three point charges are arranged along the x axis. Charge q1 = -4.40 nC is located at x= 0.210 m and charge q2 = 2.90 nC is at x=-0.330 m . A positive point charge q3 is located at the origin.

PART A: What must the value of q3 be for the net force on this point charge to have magnitude 4.40 N ?

PART B: What is the direction of the net force on q3?

PART C: Where along the x axis can q3 be placed and the net force on it be zero, other than the trivial answers of x=+ and x=?

Constants Part A Three point charges are arranged along the z axis. Charge q1 4.40 nC is located at a 0 210 mi and charge '12-2 90 nC is at z = -0 330 mA positive point charge is located at the origin. What must the value of gt be for the net force on this point charge to have magnitude 4.40 N? : nC Submit Part B What is the direction of the net force on gs? Please Choose Submit Part C Where along the x axis can gi be placed and the not force on it bo zero other than the trivial answers of z = too and z =-oo?

Explanation / Answer

a)

Net force at origin:

Fnet = (k*q1/r1^2 + k*q2/r2^2)*q3

= (9*10^9*(4.4*10^-9/0.21^2 + 2.9*10^-9/0.33^2))*q3

= 1137.6*q3

So, for Fnet = 4.4 uN,

1137.6*q3 = 4.4*10^-6

So, q3 = 3.9*10^-9 C = 3.9 nC <------ answer

b)

direction of force = towards +x axis

c)

Let the position of the charge from the charge q2 be d (towards left of q2)

So, the electric field at that point due to q2 , E1 =k*q2/d^2

Similarly, the electric field at that point due to q1, E2 = k*q1/(0.21 + 0.33 + d)^2

So, 2.9/d^2 = 4.4/(0.54 + d)^2

So, d = 2.32 m

So, the position along x axis , x = d + 0.33 = 2.65 m <------- answer

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.