Two charges are fixed on the x axis: one with a charge of q 1 = 5.00 C at x 1 =

Question

Two charges are fixed on the x axis: one with a charge of q1 = 5.00 C at x1 = -1.00 m and the other with a charge of q2 = 3.00 C at x2 = 1.50 m . Find the force F exerted on a charge q = -5.00 C placed at the origin (x=0).

Part A

What are the magnitudes of the the three charges q, q1, and q2?

Express your answer in coulombs.

Part B

Which of the following gives the proper directions for the forces on charge q due to charges q1 and q2? F1is the force on charge q due to charge q1, and F2 is the force on charge q due to charge q2. Standard coordinates are being used (i.e., "left" is the negative x direction, and "right" is the positive x direction).

Forces F1 and F2 both point to the left. Force F1 points to the left, and force F2 points to the right. Force F1 points to the right, and force F2 points to the left. Forces F1 and F2 both point to the right.

Explanation / Answer

Force on q is F = q*E

E is the electric field at origin

due to q1 is E1 = k*q1/r1^2 = (9*10^9*5*10^-6)/(1^2) = 45000 N/C towards origin

due to q2 is E2 = k*q2/r2^2 = (9*10^9*3*10^-6)/(1.5^2) = 12000 N/C towards origin

then E = E1-E2 = 45000-12000 = 33000 N/C

Force F = q*E = (-5*10^-6*33000) = -0.165 N

A) q = 5*10^-6 C

q1 = 5*10^-6 C

q2 = 3*10^-6 C

B) F1= -q*E1 and F2 = -q*E2

E1 is towrds right ..So F1 is towatrds left

E2 is towards left ..So F2 is towards Right

So correct option is Force F1 points to the left, and force F2 points to the right.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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