Chapter 17: Problem 137
A thin semi-circular ring of radius \(r\) has a positive charge \(q\) distributed uniformly over it. The net field \(\mathbf{E}\) at the centre \(O\), is (a) \(\frac{q}{4 \pi^{2} \varepsilon_{0} r^{2}} \hat{\mathbf{j}}\) (b) \(-\frac{q}{4 \pi^{2} \varepsilon_{0} r^{2}} \hat{\mathbf{j}}\) (c) \(-\frac{q}{2 \pi^{2} \varepsilon_{0} r^{2}} \hat{\mathbf{j}}\) (d) \(\frac{q}{2 \pi^{2} \varepsilon_{0} r^{2}} \hat{\mathbf{j}}\)
Short Answer
Step by step solution
Understand the system
Consider the charge element
Determine the electric field due to a charge element
Resolve into components
Calculate net vertical component
Express final result
Choose the correct option
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Charge Distribution
When a charge is
- Spread uniformly, as in our example, it means that each portion of the arc carries the same amount of charge per unit length.
- Described mathematically, this uniformity is expressed as linear charge density, denoted by the symbol \( \lambda \).
Coulomb's Law
- \( F \) is the magnitude of the force
- \( k_e = \frac{1}{4\pi\varepsilon_0} \) is Coulomb's constant
- \( q_1 \) and \( q_2 \) are the electric charges
- \( r \) is the distance between the charges
The field can then be determined at any point, like the center of our semi-circle, using this approach.
Integration of Electric Field
Here's how it works:
- Every small charge element \( dq \) creates an electric field at the point of interest.
- Each field has a radial component with magnitude given by \( dE = \frac{k_e dq}{r^2} \).
- Due to symmetry, the horizontal components cancel out when summed, simplifying the calculation to only vertical components.
- This leads to an integral of the vertical components \( dE_y = \frac{k_e q}{\pi r^2} \cos \theta \, d\theta \), resulting in the total electric field.
Linear Charge Density
- This density allows us to express a small charge element \( dq \) as \( dq = \lambda r d\theta = \frac{q}{\pi} d\theta \).
- Using linear charge density simplifies the application of Coulomb's Law over a continuous distribution of charge.
- It's important in calculating quantities like electric fields, as it provides a straightforward link between distributed charge and its physical effects.