Math Problem Statement

What is the line charge density on a long wire if a 6.7- μg particle carrying 2.2 nC describes a circular orbit about the wire with speed 320 m/s ? Express your answer in microcoulombs per meter.

Solution

To determine the line charge density (λ) on a long wire given that a charged particle describes a circular orbit around it, we'll apply the principles of electrostatics and centripetal force.

Given:

  • Mass of the particle, m=6.7μg=6.7×109kgm = 6.7 \, \mu\text{g} = 6.7 \times 10^{-9} \, \text{kg}
  • Charge of the particle, q=2.2nC=2.2×109Cq = 2.2 \, \text{nC} = 2.2 \times 10^{-9} \, \text{C}
  • Speed of the particle, v=320m/sv = 320 \, \text{m/s}

Formula:

The centripetal force required to keep the particle in circular motion is provided by the electrostatic force between the charged particle and the line charge of the wire.

  1. Centripetal force: Fcentripetal=mv2rF_{\text{centripetal}} = \frac{mv^2}{r} where rr is the radius of the circular orbit.

  2. Electrostatic force: The electric field EE due to an infinite line charge is given by: E=λ2πϵ0rE = \frac{\lambda}{2 \pi \epsilon_0 r} where λ\lambda is the line charge density, and ϵ0\epsilon_0 is the permittivity of free space (ϵ0=8.85×1012C2/Nm2\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2).

    The force on the particle is: Felectrostatic=qE=qλ2πϵ0rF_{\text{electrostatic}} = qE = q \frac{\lambda}{2 \pi \epsilon_0 r}

Since the electrostatic force provides the centripetal force, we equate these: mv2r=qλ2πϵ0r\frac{mv^2}{r} = q \frac{\lambda}{2 \pi \epsilon_0 r}

The radius rr cancels out from both sides: λ=2πϵ0mv2q\lambda = \frac{2 \pi \epsilon_0 mv^2}{q}

Plug in the values:

λ=2π(8.85×1012C2/Nm2)×(6.7×109kg)×(320m/s)22.2×109C\lambda = \frac{2 \pi (8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2) \times (6.7 \times 10^{-9} \, \text{kg}) \times (320 \, \text{m/s})^2}{2.2 \times 10^{-9} \, \text{C}}

Calculate:

λ2π×8.85×1012×6.7×109×1024002.2×109\lambda \approx \frac{2 \pi \times 8.85 \times 10^{-12} \times 6.7 \times 10^{-9} \times 102400}{2.2 \times 10^{-9}}

λ2×3.1416×8.85×6.7×102400×10212.2×109\lambda \approx \frac{2 \times 3.1416 \times 8.85 \times 6.7 \times 102400 \times 10^{-21}}{2.2 \times 10^{-9}}

λ3.706×10122.2×109\lambda \approx \frac{3.706 \times 10^{-12}}{2.2 \times 10^{-9}}

λ1.684×103C/m\lambda \approx 1.684 \times 10^{-3} \, \text{C/m}

Finally, convert to microcoulombs per meter: λ1.684μC/m\lambda \approx 1.684 \, \mu\text{C/m}

Answer:

The line charge density on the wire is approximately 1.684 μC/m.

Would you like to know more details or have any questions?

Related Questions:

  1. How would the line charge density change if the speed of the particle doubled?
  2. What would be the impact on the orbit if the mass of the particle increased?
  3. How does the radius of the orbit relate to the line charge density?
  4. What assumptions are made in using the formula for an infinite line charge?
  5. How would the line charge density be affected if the charge of the particle was negative?

Tip:

Always check the units when plugging values into formulas to ensure consistency and correct final units.

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Math Problem Analysis

Mathematical Concepts

Electrostatics
Centripetal Force
Line Charge Density

Formulas

Centripetal force formula
Electric field due to infinite line charge formula
Electrostatic force formula

Theorems

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Suitable Grade Level

Advanced High School