Math Problem Statement

Consider a ring of charge with a linear charge density 170 nC/m. The radius of the ring is 81 cm. What is the magnitude of the electric field, in newtons per coulomb, a distance 1.64 m from the center of the ring along the axis of the ring?

Solution

The problem involves calculating the magnitude of the electric field produced by a ring of charge. Let's break down the information provided:

  • Linear charge density λ=170nC/m=170×109C/m\lambda = 170 \, \text{nC/m} = 170 \times 10^{-9} \, \text{C/m}
  • Radius of the ring R=81cm=0.81mR = 81 \, \text{cm} = 0.81 \, \text{m}
  • Distance along the axis from the center of the ring to the point where the electric field is to be calculated, z=1.64mz = 1.64 \, \text{m}

Formula for the Electric Field on the Axis of a Ring of Charge:

The electric field EE along the axis of a ring of charge at a distance zz from the center is given by:

E=14πε0qz(z2+R2)3/2E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q z}{(z^2 + R^2)^{3/2}}

Where:

  • qq is the total charge on the ring
  • ε0\varepsilon_0 is the vacuum permittivity, ε0=8.854×1012C2/N\cdotpm2\varepsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2
  • zz is the distance from the center of the ring along the axis
  • RR is the radius of the ring

Steps to Solve:

  1. Total Charge on the Ring: The total charge qq on the ring can be found by multiplying the linear charge density λ\lambda by the circumference of the ring 2πR2\pi R: q=λ2πRq = \lambda \cdot 2\pi R
  2. Substitute Values into the Electric Field Formula: Once qq is calculated, substitute qq, RR, and zz into the electric field equation to get the result.

I will calculate this next.The magnitude of the electric field at a distance of 1.64m1.64 \, \text{m} from the center of the ring along its axis is approximately 2083.9N/C2083.9 \, \text{N/C}.

Would you like further details or have any questions?

Here are five related questions for deeper understanding:

  1. How does the electric field change if the distance zz increases?
  2. What happens to the electric field if the radius of the ring is doubled?
  3. How would the electric field change if the charge density were halved?
  4. What is the significance of the vacuum permittivity ε0\varepsilon_0 in this formula?
  5. How would the calculation change if the ring were not uniformly charged?

Tip: Remember that the electric field on the axis of a charged ring decreases with distance, and approaches zero as zz becomes much larger than RR.

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Math Problem Analysis

Mathematical Concepts

Electric Field
Charge Distribution
Ring of Charge

Formulas

E = (1 / 4π ε₀) * (q z) / (z² + R²)^(3/2)
q = λ * 2πR

Theorems

Coulomb's Law
Superposition Principle

Suitable Grade Level

Undergraduate