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24. Electric Force & Field; Gauss' Law

Electric Field

Physics

24. Electric Force & Field; Gauss' Law

Electric Field

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Problem 92

Textbook Question

A thin glass rod is a semicircle of radius R, Fig. 21–81. A charge is nonuniformly distributed along the semicircle with a linear charge density given by λ = λ₀ sin θ, where λ₀ is a positive constant. Point P is at the center of the semicircle. (a) Find the electric field $\vec{E}$ (magnitude and direction) at point P. [Hint: Remember sin ( -θ) = - sin θ, so the two halves of the rod are oppositely charged.] (b) Determine the acceleration (magnitude and direction) of an electron placed at point P, assuming R = 1.0 cm and λ₀ = 1.0 μC/m.

Verified step by step guidance

Step 1: Understand the problem setup. The semicircular glass rod has a nonuniform linear charge density λ = λ₀ sin θ, where λ₀ is a positive constant. The charge distribution is symmetric about the y-axis, with positive charges on the upper half and negative charges on the lower half. Point P is at the center of the semicircle, and we need to calculate the electric field at P due to this charge distribution.

Step 2: Break the semicircle into infinitesimal charge elements dq. The linear charge density λ relates to dq as dq = λ ds, where ds is the infinitesimal arc length. Since λ = λ₀ sin θ, we can write dq = λ₀ sin θ ds. The arc length ds can be expressed in terms of the radius R and the angle θ as ds = R dθ.

Step 3: Determine the electric field contribution dE→ from each infinitesimal charge element dq. The magnitude of the electric field due to dq at point P is given by dE = (1 / (4πε₀)) * (dq / R²), where ε₀ is the permittivity of free space. The direction of dE→ depends on the angle θ, and we can resolve it into x and y components: dEx = dE cos θ and dEy = dE sin θ.

Step 4: Integrate the contributions over the semicircle. For the x-component of the electric field, integrate dEx = (1 / (4πε₀)) * (λ₀ sin θ R dθ / R²) * cos θ over θ from -π/2 to π/2. Similarly, for the y-component, integrate dEy = (1 / (4πε₀)) * (λ₀ sin θ R dθ / R²) * sin θ over the same limits. Note that due to symmetry, the x-component will cancel out, leaving only the y-component.

Step 5: Use the calculated electric field to find the acceleration of the electron. The force on the electron is F→ = qE→, where q is the charge of the electron (-e). The acceleration is then a→ = F→ / m, where m is the mass of the electron. The direction of acceleration will be opposite to the direction of the electric field due to the negative charge of the electron.

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Key Concepts

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

Electric Field Due to a Continuous Charge Distribution

The electric field at a point due to a continuous charge distribution is calculated by integrating the contributions from each infinitesimal charge element. For a semicircular rod with a nonuniform linear charge density, the electric field at point P can be determined by considering the symmetry and the varying charge density along the rod.

Linear Charge Density

Linear charge density (λ) is defined as the amount of charge per unit length along a line. In this problem, the charge density varies with the angle θ, given by λ = λ₀ sin θ, which indicates that the charge is distributed nonuniformly along the semicircle, affecting the resultant electric field at point P.

Acceleration of a Charged Particle in an Electric Field

The acceleration of a charged particle, such as an electron, in an electric field is determined by Newton's second law, F = ma, where F is the force exerted by the electric field on the charge. The force can be calculated using F = qE, where q is the charge of the particle and E is the electric field strength. This relationship allows us to find the acceleration of the electron at point P.

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Related Practice

Textbook Question

Three very large square charged planes are arranged as shown (on edge) in Fig. 21–78. From left to right, the planes have charge densities per unit area of -0.50μC/m², +0.25 μC/m² and -0.35 μC/m². Find the total electric field (direction and magnitude) at the points A, B, C, and D. Assume the plates are much larger than the distance AD.

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Textbook Question

(II) Determine the direction and magnitude of the electric field at the point P shown in Fig. 21–66. The two point charges are separated by a distance of 2a. Point P is on the perpendicular bisector of the line joining the charges, a distance x from the midpoint between them. Express your answers in terms of Q, x, a, and k.

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Textbook Question

(III) A thin rod of length ℓ carries a total charge Q distributed uniformly along its length. See Fig. 21–69. Determine the electric field along the axis of the rod starting at one end—that is, find E(𝓍) for 𝓍 ≥ 0 in Fig. 21–69.

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Textbook Question

(II) A very long uniformly charged wire (linear charge density λ = 2.5 C/m) lies along the x-axis in Fig. 21–59. A small charged sphere (Q = -2.0 C) is at the point x = 0 cm, y = -5.0 cm. What is the electric field at the point x = 7.0 cm, y = 7.0 cm? ${\vec{E}}_{wire}$ and ${\vec{E}}_{q}$ represent fields due to the long wire and the charge Q, respectively.