https://file.aiquickdraw.com/mathbot/file/1728593055281tef0zykf.jpg

The image provided shows a problem related to the electric field created by a charged arc of glass. Let's go through the key details:

### Problem Summary:
- A piece of glass is shaped like an arc with a radius of \( R = \frac{4}{3} \) meters.
- The arc subtends an angle \( 2\alpha \), where \( \alpha = \frac{\pi}{3} \) radians.
- The arc carries a charge density \( \lambda = 4 \, \mu C/m \) (charge per unit length).
- The goal is to find the electric field at point \( O \), which is located at the center of curvature of the arc.

Given:
- Coulomb constant \( k = 9000 \, \text{Vm}/\mu\text{C} \).
- You are asked to write the definite integral to compute the electric field strength and evaluate it, approximating the field strength at a distance of 3 km from the arc.

### Part (a): Defining the Electric Field Integral
- The electric field at the center due to a small element of charge \( dq \) is derived from Coulomb's law:
  \[
  dE = \frac{k \, dq}{R^2}
  \]
- Since the charge is distributed along the arc, \( dq \) is related to the charge density \( \lambda \) and the arc length element \( ds \), which is expressed as \( R d\theta \):
  \[
  dq = \lambda R d\theta
  \]
- Substituting \( dq \) into the electric field formula:
  \[
  dE = \frac{k \, \lambda \, R \, d\theta}{R^2}
  \]
  Simplifying:
  \[
  dE = \frac{k \, \lambda \, d\theta}{R}
  \]
- The total electric field is the integral of \( dE \) over the angle \( \theta \), from \( -\alpha \) to \( \alpha \) (the limits of the arc):
  \[
  E = \int_{-\alpha}^{\alpha} \frac{k \, \lambda \, d\theta}{R}
  \]
- With the given values \( \alpha = \frac{\pi}{3} \), \( k = 9000 \, \text{Vm}/\mu\text{C} \), \( \lambda = 4 \, \mu\text{C}/m \), and \( R = \frac{4}{3} \), the expression for the electric field becomes:
  \[
  E = \frac{k \, \lambda}{R} \int_{-\alpha}^{\alpha} d\theta
  \]
  Since \( \int_{-\alpha}^{\alpha} d\theta = 2\alpha \), we have:
  \[
  E = \frac{9000 \times 4}{\frac{4}{3}} \times 2 \times \frac{\pi}{3}
  \]
  Simplifying:
  \[
  E = \frac{36000}{\frac{4}{3}} \times \frac{2\pi}{3} = 27000 \times \frac{2\pi}{3}
  \]
  Finally:
  \[
  E = 18000 \pi \, \text{Vm}/m
  \]

### Part (b): Evaluating the Electric Field
Approximating the value:
\[
E \approx 18000 \times 3.1416 = 56548.67 \, \text{Vm}/m
\]

### Part (c): Approximation at 3 km
The problem asks to estimate the electric field strength at a distance of 3 kilometers. Since this is far compared to the arc radius, the electric field strength would significantly decrease due to the inverse square law. However, from this image, there's no clear guidance on how this part connects, so we can assume that the field might weaken dramatically.

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Let me know if you want further explanations on any step or clarification on other parts. 

Here are some relative questions:
1. How would the electric field change if the charge density doubled?
2. What would happen to the electric field if the arc radius increased?
3. How would you compute the electric field if the charge distribution were non-uniform?
4. How does the electric field behave at very far distances compared to near the arc?
5. How does the angle subtended by the arc affect the electric field at the center?

**Tip:** Always check the units of each physical quantity to ensure consistency when setting up integrals.

A piece of glass is shaped like an arc with a curvature radius of 4/3 meters. The arc subtends an angle of 2α, where α = π/3 radians, and carries a charge density of 4 μC/m. The goal is to compute the electric field strength at point O, located at the center of curvature of the arc.

Solve the electric field problem for an arc-shaped glass with a radius of 4/3 meters and a charge density of 4 μC/m. Use integral calculus and Coulomb's law to find the electric field at the center of the arc's curvature. Ideal for undergraduate electromagnetism studies.

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