Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

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Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

and the other particle is at x = 100.00 cm. Where on the x-axis must a third charged particle be placed so that it does not experience any net electrostatic force due to the other two particles? Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm, and the other particle is at x = 100.00 cm. Where on the x-axis must a third charged particle be placed so that it does not experience any net electrostatic force due to the other two particles? 4.58 cm 50 cm 9.12 cm 91.2 cm 45.8 cm

Physics

1 answer:

Sav [3.1K] · Answer 1 · 2026-03-09T05:04:10+03:00

Answer:

The third charged particle needs to be positioned at x = 0.458 m = 45.8 cm

Explanation:

To address this inquiry, we refer to Coulomb's law:

Two point charges (q₁, q₂) spaced apart by a distance (d) impart a mutual force (F) with a magnitude defined by the equation:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K: Coulomb's constant in N*m²/C²

q₁, q₂: Charges quantified in Coulombs (C)

d: distance in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Information

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem analysis

Refer to the accompanying graphic.

We assume a positive charge q₃, thus F₁₃ and F₂₃ exhibit repulsive forces that need to be equivalent so the resultant force is zero:

Utilizing equation (1), we will determine the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ from both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We discard 10⁻⁶ from both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve for the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

Choosing x₂ results in F₁₃ and F₂₃ acting in the same direction and unable to cancel each other, therefore we select x₁ as the accurate choice since at this stage the forces counteract each other.

x = 0.458m = 45.8cm