A charge q1=6
𝜇
C is at position (x1,y1), another charge q2=9
𝜇
C is at position (x2,y2) and a third charge q3=2
𝜇
C is at position (x3,y3). Find the x component of the total force exerted by q1 and q2 on q3. The unit of your answer is N. (
𝑘
=
1
(
4
𝜋
𝜖
𝑜
)
=8.99x109Nm2/C2)

Assume

x1=6cm, y1=6cm

x2=4cm, y2=7cm

x3=4cm,y3=8cm

Question

A charge q1=6
𝜇
C is at position (x1,y1), another charge q2=9
𝜇
C is at position (x2,y2) and a third charge q3=2
𝜇
C is at position (x3,y3). Find the x component of the total force exerted by q1 and q2 on q3. The unit of your answer is N. (
𝑘
=
1
(
4
𝜋
𝜖
𝑜
)
=8.99x109Nm2/C2)

Assume

x1=6cm, y1=6cm

x2=4cm, y2=7cm

x3=4cm,y3=8cm

BlackTom AI · Accepted Answer

We start by listing the given quantities and converting units to be consistent. The charges are q1 = 6 μC, q2 = 9 μC, q3 = 2 μC. In coulombs: q1 = 6e-6 C, q2 = 9e-6 C, q3 = 2e-6 C. The positions in meters are: r1 = (0.06, 0.06), r2 = (0.04, 0.07), r3 = (0.04, 0.08).

For each charge qi, the force on q3 due to qi is given by F_i = k * q3 * qi * (r3 − ri) / |r3 − ri|^3, where k = 8.99e9 N·m^2/C^2.

1) Force from q1:
- Vector from q1 to q3: r31 = r3 − r1 = (0.04 − 0.06, 0.08 − 0.06) = (−0.02, 0.02).
- Distance |r31| = sqrt((−0.02)^2 + (0.02)^2) = sqrt(0.0004 + 0.0004) = sqrt(0.0008) ≈ 0.028284 m.
- Cubed distance: |r31|^3 ≈ (0.028284)^3 ≈ 2.2627e−5 m^3.
- Product q3 * q1 = (2e−6)(6e−6) = 12e−12 = 1.2e−11 C^2.
- Constant factor for this interaction: C1 = k * (q3 * q1) / |r31|^3 ≈ (8.99e9 * 1.2e−11) / 2.2627e−5 ≈ 0.10788 / 2.2627e−5 ≈ 4.77e3 N/m.
- Components of F1: F1 = C1 * r31 ≈ 4.77e3 * (−0.02, 0.02) ≈ (−95.3, +95.3) N.

2) Force from q2:
- Vector from q2 to q3: r32 = r3 − r2 = (0.04 − 0.04, 0.08 − 0.07) = (0, 0.01).
- Distance |r32| = sqrt(0^2 + 0.01^2) = 0.01 m. |r32|^3 = (0.01)^3 = 1e−6 m^3.
- Product q3 * q2 = (2e−6)(9e−6) = 18e−12 = 1.8e−11 C^2.
- Constant factor: C2 = k * (q3 * q2) / |r32|^3 = (8.99e9 * 1.8e−11) / 1e−6 = 0.16182 / 1e−6 ≈ 1.6182e5 N/m.
- Components of F2: F2 = C2 * r32 = 1.6182e5 * (0, 0.01) = (0, 1618.2) N.

Now sum the x-components to get the total x-component of the force on q3 due to q1 and q2:
- F_x,total = F1_x + F2_x = (−95.3) + 0 ≈ −95.3 N.

The computed total x-component is approximately −95.3 N. Depending on rounding at intermediate steps, this value may appear as about −95.5 N in a rounded result. The key idea is that the x-contribution comes from the x-component of the force from q1 (which has an equal negative x and positive y component due to the geometry) plus the x-component from q2 (which lies directly above q3, contributing no x-component).

202502PHY1202 Assignment 2 Electric charges - due February 9, 2025, 11:59pm

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