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Difficulty: Very hardElectrostatics and Electric Charges

Two identical point charges, q1=+5.0×10−6 Cq_1 = +5.0 \times 10^{-6}\text{ C} and q2=+5.0×10−6 Cq_2 = +5.0 \times 10^{-6}\text{ C}, are fixed in a vacuum at Cartesian coordinates (0 m,3.0 m)(0\text{ m}, 3.0\text{ m}) and (0 m,−3.0 m)(0\text{ m}, -3.0\text{ m}), respectively. A third point charge q3=+2.0×10−6 Cq_3 = +2.0 \times 10^{-6}\text{ C} is placed on the x-axis at (4.0 m,0 m)(4.0\text{ m}, 0\text{ m}). Taking Coulomb's constant k=9.0×109 N m2 C−2k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}, what is the magnitude of the net electrostatic force exerted on q3q_3?

  1. 5.76×10−3 N5.76 \times 10^{-3}\text{ N}Answer
  2. B
    7.20×10−3 N7.20 \times 10^{-3}\text{ N}
  3. C
    4.32×10−3 N4.32 \times 10^{-3}\text{ N}
  4. D
    3.60×10−3 N3.60 \times 10^{-3}\text{ N}

Answer

The magnitude of the net electrostatic force exerted on q3q_3 is 5.76×10−3 N5.76 \times 10^{-3}\text{ N}.
Each charge exerts a repulsive force of magnitude 3.60×10−3 N3.60 \times 10^{-3}\text{ N} along the line connecting it to the test charge at (4.0 m,0 m)(4.0\text{ m}, 0\text{ m}). Because of symmetry, the y-components of the two forces cancel out completely while their x-components add constructively. Multiplying the individual force magnitude by the cosine of the angle with the x-axis (cos⁡θ=0.8\cos\theta = 0.8) and doubling for both charges gives a net force of 5.76×10−3 N5.76 \times 10^{-3}\text{ N}.

Step-by-Step Solution

1
Calculate the straight-line distance rr from q1q_1 (or q2q_2) to q3q_3.
r=(4.0−0)2+(0−3.0)2=16+9=5.0 mr = \sqrt{(4.0 - 0)^2 + (0 - 3.0)^2} = \sqrt{16 + 9} = 5.0\text{ m}.
Coulomb's Law requires the straight-line separation distance between interacting point charges.
2
Calculate the magnitude of the electrostatic force F1F_1 exerted on q3q_3 by q1q_1.
F1=k⋅q1⋅q3r2=(9.0×109)⋅(5.0×10−6)⋅(2.0×10−6)5.02=9.0×10−225=3.60×10−3 NF_1 = \frac{k \cdot q_1 \cdot q_3}{r^2} = \frac{(9.0 \times 10^9) \cdot (5.0 \times 10^{-6}) \cdot (2.0 \times 10^{-6})}{5.0^2} = \frac{9.0 \times 10^{-2}}{25} = 3.60 \times 10^{-3}\text{ N}.
By symmetry, the force magnitude F2F_2 from q2q_2 on q3q_3 is also equal to 3.60×10−3 N3.60 \times 10^{-3}\text{ N}.
3
Determine the vector components of the forces along the axes.
The cosine of the angle θ\theta with the positive x-axis is cos⁡θ=4.05.0=0.8\cos\theta = \frac{4.0}{5.0} = 0.8. The sine is sin⁡θ=3.05.0=0.6\sin\theta = \frac{3.0}{5.0} = 0.6. The vertical y-components are equal in magnitude and opposite in direction (F1y=−F2yF_{1y} = -F_{2y}), canceling to zero.
Forces are vector quantities; symmetrically placed identical charges produce opposing vertical components and reinforcing horizontal components.
4
Sum the horizontal x-components to obtain the net force.
Fnet=F1x+F2x=2⋅F1cos⁡θ=2⋅(3.60×10−3 N)⋅0.8=5.76×10−3 NF_{\text{net}} = F_{1x} + F_{2x} = 2 \cdot F_1 \cos\theta = 2 \cdot (3.60 \times 10^{-3}\text{ N}) \cdot 0.8 = 5.76 \times 10^{-3}\text{ N}.
Both x-components point along the positive x-axis, so their magnitudes add directly.

Key Concept

Vector superposition of electric forces
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Topics

Electrostatics and Electric Charges

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