Three charged particles are placed at the comers of an equilateral triangle of s

Question

Three charged particles are placed at the comers of an equilateral triangle of side a = 1.20 m, as shown in the figure. The charges are: q1 = 7.00 mu C, q2 = -8.00 mu C, and q3 = -6.00 mu C. Calculate the following: The magnitude F12 of the force exerted on q1 by q2. Draw on the figure. The magnitude F13 of the force exerted on q1 by q3. Draw on the figure. The x-component F1x of the net force exerted on q1 by q2and q3. The y-component F1y of the net force exerted on q1 by q2 and q3. The magnitude F1 of the net force . The angle Theta between and the x-axis.

Explanation / Answer

On converting, µC to C, we get

Q1 = 7x10-6 C

Q2 = -8x10-6 C

Q3 = -6x10-6 C

You'll notice the signs of the forces - both F and F are negative, and only the x-component of the F force is positive. Also, 60° because it's an equilateral triangle.

F = k*Q2*Q1/1.3²

F = (9.0 x 109) * (-8x10-6) * (7x10-6)/ 1.3² = -0.29882 N

2. Magnitude of force on Q1 due to Q3 =

F = k*Q1*Q3/1.3²

F = (9.0 x 109) * (-6x10-6) * (7x10-6)/ 1.3² = -0.2236 N

3. Now you must add up the components in vector form:
F1x = Fcos(60°) - Fcos(60°)

          F1x = ((-0.29882) * cos60) – ((-0.2236) * cos60) = -0.03761 N

4. F1y = Fsin(60°) + Fsin(60°)

Perform the same calculations as in part no (c).

5. Now we know that,

F = k*Q3*Q1/1.3²
F = k*Q3*Q2/1.3²

Also,

Fx = Fcos(60°) + F
Fy = -Fsin(60°)

Therefore, to determine magnitude of the net force on a particle, F = sqrt(Fx²+Fy²).

6. Now to find the angle;

Use the below mentioned formula =arctan(Fy/Fx).

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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