1.5. Study the following problem that was solved directly from Coulomb's law and answer the set of questions that follow. • Problem Calculate the electric field at a point P created by a thin, long, straight filament, electrically charged with a constant linear density A, using the direct approach from Coulomb's law.
1.5. Study the following problem that was solved directly from Coulomb's law and answer the set of questions that follow. • Problem Calculate the electric field at a point P created by a thin, long, straight filament, electrically charged with a constant linear density A, using the direct approach from Coulomb's law.
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Question
Transcribed Image Text:2=+00
2=-00
dE
Then, according to the linear superposition principle, the field's magnitude may
be calculated as
E(p) =
so that the total field
2p
E(p) =
P
Questions
dE cose
+00
(p² +2²) ¹/2
•=+00
d
dEp = de cos 0 =
where dE is the magnitude of the elementary
contribution to the field, created by a small segment dz
of the filament, with electric charge Adz.
But
dz
E
4лE-0 (p²+²)³/2 -
Z=+00
dE=Adz-
-00
2=-0
1 1
Απευ ρ' + 22'
dE-
P
(p² +2²) ¹/2²
¹+00
2лεo Jo
de
λ
(1 +52)³/2 Σπειρ
=
1.5.1. Assume that the person who solved this problem
was a candidate writing an examination. Why did
the candidate considere using the translational and
axial symmetries of the
problem?
1.5.2. The candidate used the principle of superposition
as part of his reasoning the solution. State the
principle of superposition in words in terms of the
electric field
vector
1.5.3. Write down the mathematical expression from the
candidate solution, that represents the principle of
superposition
1.5.4. Why did the candidate chose both the lower and
upper limit of the integral as - and +
?
1.5.5. Evaluate the candidate's answer giving reasons
whether it makes sense or
not.
Transcribed Image Text:1.5. Study the following problem that was solved directly
from Coulomb's law and answer the set of questions
that follow.
• Problem
Calculate the electric field at a point P created by a
thin, long, straight filament, electrically charged with a
constant linear density A, using the direct approach
from Coulomb's law.
Solution
(i) From the translational and axial symmetries of the problem, it is clear that E(r) =
nE(p), where p is the shortest distance from the observation point to the filament'.
Let us select the plane of drawing so that it contains both the filament and the
observation point, and take the line of the filament for axis z (see the figure below).
dE cos 0
2=+00
E(p) =
S
Then, according to the linear superposition principle, the field's magnitude may
be calculated as
Z=-00
dE
2=+00
dEp = f* de cos 0 = = f*t* dE=
P
Z=-00
(p² +2²)¹/2
where dE is the magnitude of the elementary
contribution to the field, created by a small segment dz
of the filament, with electric charge Adz.
But
so that the total field
0
+00
E(p) = AP fo
P
Questions
(P² +2²)¹/2
dz
4лE-00 (p² +2²)3/2
1 1
4лeo p² +₂²³
dE=Adz-
E
λ
de
Σπερίο (1+2) 3/2 Σπειρ
·+·00
1.5.1. Assume that the person who solved this problem
was a candidate writing an examination. Why did
the candidate considere using the translational and
axial symmetries of the
problem?
1.5.2. The candidate used the principle of superposition
as part of his reasoning the solution. State the
principle of superposition in words in terms of the
electric field
vector
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