Question

There is a uniform charge distribution of \(\lambda=6.005 \cdot 10^{-8} \mathrm{C} / \mathrm{m}\) along a thin wire of length \(L .\) The wire is then curved into a semicircle that is centered at the origin and has a radius of \(R=L / \pi .\) The magnitude of the electric field at the center of the semicircle is \(2.425 \cdot 10^{4} \mathrm{~N} / \mathrm{C}\). What is the value of \(L ?\)

Short answer

Question: Given a thin wire curved into a semicircle with a uniform charge distribution, a radius R equal to L / π, a charge density λ, and an electric field magnitude at the center, find the length L of the wire. Answer: The length L of the wire is approximately 6.50 meters.

Step-by-step solution

Find the total charge

The charge, Q, is given by the charge density multiplied by the length of the wire. We can represent this as: $$ Q = \lambda L $$

Use expression for the electric field at the center of a semicircle

The electric field, E, at the center of a semicircle due to its charge is given by the formula: $$ E = \frac{Q}{4 \pi \varepsilon_0 R} $$ where \(\varepsilon_0\) is the vacuum permittivity, with a value of \(8.854 \times 10^{-12} \mathrm{~C^2 / N m^2}\).

Substitute expressions for Q and R into the electric field formula

Use the expressions for Q and R from steps 1 and 2 to rewrite the electric field formula as: $$ E = \frac{\lambda L}{4 \pi \varepsilon_0 (L / \pi)} $$

Solve for L

We are given the value of E as \(2.425 \times 10^4 \mathrm{N / C}\), and the value of λ as \(6.005 \times - 10^{-8} \mathrm{C / m}\). We want to find L. Substitute the given values into the formula obtained in step 3, and solve for L: $$ 2.425 \times 10^4 = \frac{(6.005 \cdot 10^{-8})(L)}{4 \pi (8.854 \times 10^{-12})(L / \pi)} $$ By solving for L, we get: $$ L = \frac{2.425 \times 10^4 \cdot 4 \pi (8.854 \times 10^{-12})}{6.005 \cdot 10^{-8}} $$ $$ L \approx 6.50~\mathrm{m} $$ So, the value of L is approximately 6.50 meters.