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Ch 23: The Electric Field
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 23: The Electric FieldProblem 48a
Chapter 23, Problem 48a

Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in FIGURE P23.48. Find an expression for the electric field Ē at the center of the semicircle. Hint: A small piece of arc length Δs spans a small angle Δθ=Δs/R , where R is the radius.
A semicircular rod with uniform charge distribution, showing the center point and arc length labeled.

Verified step by step guidance
1
Start by understanding the problem: The charge Q is uniformly distributed along the length of the rod, which is bent into a semicircle of radius R. The goal is to find the electric field at the center of the semicircle due to this charge distribution.
Divide the semicircle into small elements of charge dq. Since the charge is uniformly distributed, the linear charge density λ is given by λ = Q / L, where L is the length of the semicircle. For a semicircle, L = πR, so λ = Q / (πR).
Consider a small arc length Δs on the semicircle. The charge on this small segment is dq = λΔs. Using the hint, Δs = RΔθ, so dq = λRΔθ.
The electric field contribution dĒ from this small charge element dq at the center of the semicircle has a magnitude given by Coulomb's law: dE = (1 / (4πε₀)) * (dq / R²). Substituting dq = λRΔθ, we get dE = (1 / (4πε₀)) * (λRΔθ / R²).
Since the semicircle is symmetric, the horizontal components of the electric field from opposite sides cancel out, leaving only the vertical components. The vertical component of dĒ is dE_y = dE * sin(θ). Integrate dE_y over the angle θ from 0 to π to find the total electric field at the center: E_y = ∫[0 to π] (1 / (4πε₀)) * (λR / R²) * sin(θ) dθ. Simplify and solve the integral to express the electric field in terms of Q, R, and fundamental constants.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

The electric field is a vector field that represents the force exerted by an electric charge on other charges in its vicinity. It is defined as the force per unit charge and is measured in newtons per coulomb (N/C). The direction of the electric field is away from positive charges and towards negative charges, and it can be calculated using Coulomb's law or by integrating contributions from continuous charge distributions.
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Continuous Charge Distribution

A continuous charge distribution refers to a scenario where charge is spread out over a certain length, area, or volume, rather than being concentrated at discrete points. In this case, the charge is uniformly distributed along the length of the rod. To analyze the electric field from such distributions, we often use calculus to integrate the contributions from infinitesimally small charge elements.
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Arc Length and Angle Relationship

In circular motion, the relationship between arc length (Δs), radius (R), and the subtended angle (Δθ) is given by the formula Δs = RΔθ. This relationship is crucial for understanding how small segments of a circular arc contribute to the overall electric field at a point, such as the center of a semicircle. By relating these quantities, we can express the contributions of each small charge element in terms of the angle it subtends.
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Related Practice
Textbook Question

The two parallel plates in FIGURE P23.53 are 2.0 cm apart and the electric field strength between them is 1.0×104 N/C. An electron is launched at a 45° angle from the positive plate. What is the maximum initial speed v0 the electron can have without hitting the negative plate?

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Textbook Question

CALC A uniform electric field’s strength is increasing with time as E=(1.5×104+(5.0×1010s−1)t)N/CE = (1.5 \(\times\) 10^4 + (5.0 \(\times\) 10^{10}\,\(\text{s}\)^{-1})t)\,\(\text{N/C}\). A proton is released in the field from rest at t = 0. What is the proton’s speed 1.0 μs later?

Textbook Question

Derive Equation 23.11 for the field Ē dipole in the plane that bisects an electric dipole.

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Textbook Question

FIGURE P23.44 shows a thin rod of length L with total charge Q. Evaluate E at r=3.0 cm if L=5.0 cm and Q=3.0 nC.

Textbook Question

An infinite plane of charge with surface charge density 3.2 μC/m2 has a 20-cm-diameter circular hole cut out of it. What is the electric field strength directly over the center of the hole at a distance of 12 cm? Hint: Can you create this charge distribution as a superposition of charge distributions for which you know the electric field?

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Textbook Question

A ring of radius R has total charge Q. At what distance along the z-axis is the electric field strength a maximum?