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答案：
第1步
Given:
Center@origininthez = 0
plane
Inner radius =A
Outer radius =B
T heunif ormsurf acechargerhos (C/m²) = (rhoo , A ≤ rho ≤ B), (0, otherwise) :
Concept:
Cylindrical coordinates are a coordinate system used in three-dimensional space to describe the position of a point. They consist of three parameters: radial
distance (ρ), azimuthal angle (φ), and vertical position (z).
The radial distance (ρ) represents the distance from the origin to the point in a direction perpendicular to the z-axis.
The azimuthal angle (φ) measures the angle formed by a line connecting the origin to the point with respect to a reference axis, typically the x-axis.
The vertical position (z) indicates the distance of the point from the z-axis, with positive values above the xy-plane and negative values below it.
To find the electric field vector above the center of the ring, we can use the principle of superposition. We'll divide the ring into infinitesimally small charge ele
and integrate over the entire ring.
第2步
1. Finding vector E (ρ = 0, φ, z) above the center of the ring:
Since the ring is very thin, we can consider it as a line charge. Each infinitesimally small charge element on the ring will contribute to the electric field at poin
0, φ, z) in the upward direction.
Let's denote the linear charge density as λ = ρo, where ρo is the uniform surface charge density. The charge element on the ring, dQ, is given by dQ = λ dφ,
dφ is the differential angle subtended by the charge element.
F oreachchargeelement, risgivenby : r = √(z
2
2
+ ρ )
Now, we can write the electric field vector at point P as a superposition of the contributions from all charge elements on the ring:
E(ρ = 0, φ, z) = ∫
dE
Integrating this expression, we get:
E(ρ = 0, φ, z) = ∫ (kλdφ)/(z
2
2
(
+ ρ ) 3/2)
The integration limits will be from 0 to 2π since we're considering a complete ring.
第3步
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9/14/23, 3:30 PM
Chegg 24-hour self-service query answers coursehero
Replacing ρdφ with du/2 and simplifying, we have:
(
E(ρ = 0, φ, z) = ∫ (kλ/2)(du/u 3/2))
Evaluating the integral, we get:
evaluatedf roA
2
2
+ z toB
2
+ z
2
m
Simplifying further, we have:
E(ρ = 0, φ, z) = −(kλ/2)[(B
2
2
(
2
(
+ z )
− 1/2) − (A
2
2
(
2
(
+ z )
− 1/2)]
Therefore, the electric field vector above the center of the ring is:
E(ρ = 0, φ, z) = −(kλ/2)[(B
2
+ z )
− 1/2) − (A
2
+ z )
− 1/2)]
第4步
2. Determining the relationship between E(ρ = 0, φ, -z) and E(ρ = 0, φ, z):
To determine if E(ρ = 0, φ, -z) is parallel or antiparallel to E(ρ = 0, φ, z), we can compare their directions.
F orE(ρ = 0, φ, z)
, the electric field points in the upward direction.
F orE(ρ = 0, φ, −z)
, the electric field will also point in the upward direction since the distance from the charge elements on the ring to the point is the same as in the previous c
T hus, isparalleltE(ρ = 0, φ, z)
o .
The electric field vector at a point is a measure of the force experienced by a positive test charge placed at that point due to surrounding charges. It is define
the force per unit positive charge.
Final answer
1.
E(ρ = 0, φ, z) = −(kλ/2)[(B
2.
2
2
+ z )
(
− 1/2) − (A
2
2
+ z )
(
− 1/2)]
is parallel to .
E(ρ = 0, φ, −z)
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