Question

Difficulty: Very hardElectrostatics and Electric Charges

Two identical positive point charges, each of magnitude q=+2.5×106 Cq = +2.5 \times 10^{-6}\text{ C}, are fixed in a vacuum at a distance of 0.60 m0.60\text{ m} apart. A third point charge q0=+1.0×106 Cq_0 = +1.0 \times 10^{-6}\text{ C} is placed on the perpendicular bisector of the line joining the two fixed charges, at a distance of 0.40 m0.40\text{ m} from their midpoint. Taking Coulomb's constant k=9.0×109 N m2 C2k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}, what is the magnitude of the net electrostatic force acting on the third charge, in newtons?

Answer: 0.144 N

Answer

The magnitude of the net electrostatic force acting on the third charge is 0.144 N0.144\text{ N}.
Each fixed charge exerts an equal repulsive electrostatic force of 0.09 N0.09\text{ N} on the third charge. Due to the symmetrical arrangement, the force components perpendicular to the bisector cancel each other out, while the parallel components add together, giving a net force of 2×0.09×0.8=0.144 N2 \times 0.09 \times 0.8 = 0.144\text{ N}.

Step-by-Step Solution

1
Determine the distance from each fixed charge to the third charge
r=0.50 mr = 0.50\text{ m}
The charges form a right-angled triangle with base 0.30 m0.30\text{ m} (half of 0.60 m0.60\text{ m}) and height 0.40 m0.40\text{ m}, yielding a hypotenuse of 0.302+0.402=0.50 m\sqrt{0.30^2 + 0.40^2} = 0.50\text{ m}.
2
Calculate the magnitude of the individual repulsive force from one charge
F=0.09 NF = 0.09\text{ N}
Applying Coulomb's law: F=kqq0r2=9.0×109×2.5×106×1.0×1060.25=0.09 NF = \frac{k q q_0}{r^2} = \frac{9.0 \times 10^9 \times 2.5 \times 10^{-6} \times 1.0 \times 10^{-6}}{0.25} = 0.09\text{ N}.
3
Determine directional component of forces along the perpendicular bisector
cosθ=0.8\cos\theta = 0.8
The directional cosine along the axis of symmetry is the ratio of the adjacent side (0.40 m0.40\text{ m}) to the hypotenuse (0.50 m0.50\text{ m}).
4
Compute the net electrostatic force using vector addition
Fnet=0.144 NF_{\text{net}} = 0.144\text{ N}
Horizontal components cancel by symmetry, so Fnet=2Fcosθ=2×0.09×0.8=0.144 NF_{\text{net}} = 2 F \cos\theta = 2 \times 0.09 \times 0.8 = 0.144\text{ N}.

Key Concept

Vector superposition of Coulombic forces along an axis of symmetry
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