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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: 9780137344796당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Knight Calc 5th EditionCh 23: The Electric Field문제 50
23장, 문제 50

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?

검증된 단계별 안내
1
Recognize that the problem involves a superposition principle. The electric field at the center of the hole can be calculated by considering the infinite plane of charge (with surface charge density \( \sigma = 3.2 \mu C/m^2 \)) and subtracting the contribution of the circular hole (treated as a disk of charge with the same surface charge density).
Write the expression for the electric field due to an infinite plane of charge. The electric field produced by an infinite plane of charge is uniform and given by \( E_{plane} = \frac{\sigma}{2\varepsilon_0} \), where \( \varepsilon_0 \) is the permittivity of free space.
Determine the electric field due to the circular hole. Treat the hole as a disk of charge with radius \( r = 0.1 \) m (since the diameter is 20 cm). The electric field at a distance \( z = 0.12 \) m above the center of a uniformly charged disk is given by \( E_{disk} = \frac{\sigma}{2\varepsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + r^2}} \right) \).
Apply the superposition principle. The total electric field at the point directly above the center of the hole is \( E_{total} = E_{plane} - E_{disk} \). Substitute the expressions for \( E_{plane} \) and \( E_{disk} \) into this equation.
Simplify the expression for \( E_{total} \) by substituting the given values: \( \sigma = 3.2 \mu C/m^2 \), \( z = 0.12 \) m, \( r = 0.1 \) m, and \( \varepsilon_0 = 8.85 \times 10^{-12} \) C^2/(N·m^2). This will yield the electric field strength directly over the center of the hole.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Electric Field Due to a Plane of Charge

An infinite plane of charge creates a uniform electric field that is perpendicular to its surface. The strength of this electric field (E) is given by the formula E = σ / (2ε₀), where σ is the surface charge density and ε₀ is the permittivity of free space. This concept is crucial for understanding how the electric field behaves in the presence of a charged plane.
추천 영상:
가이드 코스
06:28
Electric Field due to a Point Charge

Superposition Principle

The superposition principle states that the total electric field created by multiple charge distributions is the vector sum of the electric fields produced by each distribution individually. In this problem, the infinite plane of charge can be thought of as the sum of the field from the full plane and the field from a circular hole, allowing for easier calculation of the resultant field at the specified point.
추천 영상:
가이드 코스
03:32
Superposition of Sinusoidal Wave Functions

Electric Field of a Circular Hole

To find the electric field at the center of a circular hole in an infinite plane, one can consider the hole as a negative charge distribution that cancels out part of the field from the plane. The electric field due to a circular disk can be calculated using integration techniques, and its effect can be subtracted from the field of the full plane to determine the net electric field at the center of the hole.
추천 영상:
가이드 코스
5:24
Black Holes
관련 실천
교과서 질문

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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교과서 질문

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?

1113
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교과서 질문

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.

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교과서 질문

A problem of practical interest is to make a beam of electrons turn a 90° corner. This can be done with the parallel-plate capacitor shown in FIGURE P23.55. An electron with kinetic energy 3.0×10−17 J enters through a small hole in the bottom plate of the capacitor. Should the bottom plate be charged positive or negative relative to the top plate if you want the electron to turn to the right? Explain.

229
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교과서 질문

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.

73
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교과서 질문

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

49
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