The increased availability of biomedical data, particularly in the public domain, offers the opportunity to better understand human health and to develop effective therapeutics for a wide range of unmet medical needs. However, data scientists remain stymied by the fact that data remain hard to find and to productively reuse because data and their metadata i) are wholly inaccessible, ii) are in non-standard or incompatible representations, iii) do not conform to community standards, and iv) have unclear or highly restricted terms and conditions that preclude legitimate reuse. These limitations require a rethink on data can be made machine and AI-ready - the key motivation behind the FAIR Guiding Principles. Concurrently, while recent efforts have explored the use of deep learning to fuse disparate data into predictive models for a wide range of biomedical applications, these models often fail even when the correct answer is already known, and fail to explain individual predictions in terms that data scientists can appreciate. These limitations suggest that new methods to produce practical artificial intelligence are still needed.
In this talk, I will discuss our work in (1) building an integrative knowledge infrastructure to prepare FAIR and "AI-ready" data and services along with (2) neurosymbolic AI methods to improve the quality of predictions and to generate plausible explanations. Attention is given to standards, platforms, and methods to wrangle knowledge into simple, but effective semantic and latent representations, and to make these available into standards-compliant and discoverable interfaces that can be used in model building, validation, and explanation. Our work, and those of others in the field, creates a baseline for building trustworthy and easy to deploy AI models in biomedicine.
Bio
Dr. Michel Dumontier is the Distinguished Professor of Data Science at Maastricht University, founder and executive director of the Institute of Data Science, and co-founder of the FAIR (Findable, Accessible, Interoperable and Reusable) data principles. His research explores socio-technological approaches for responsible discovery science, which includes collaborative multi-modal knowledge graphs, privacy-preserving distributed data mining, and AI methods for drug discovery and personalized medicine. His work is supported through the Dutch National Research Agenda, the Netherlands Organisation for Scientific Research, Horizon Europe, the European Open Science Cloud, the US National Institutes of Health, and a Marie-Curie Innovative Training Network. He is the editor-in-chief for the journal Data Science and is internationally recognized for his contributions in bioinformatics, biomedical informatics, and semantic technologies including ontologies and linked data.
The increased availability of biomedical data, particularly in the public domain, offers the opportunity to better understand human health and to develop effective therapeutics for a wide range of unmet medical needs. However, data scientists remain stymied by the fact that data remain hard to find and to productively reuse because data and their metadata i) are wholly inaccessible, ii) are in non-standard or incompatible representations, iii) do not conform to community standards, and iv) have unclear or highly restricted terms and conditions that preclude legitimate reuse. These limitations require a rethink on data can be made machine and AI-ready - the key motivation behind the FAIR Guiding Principles. Concurrently, while recent efforts have explored the use of deep learning to fuse disparate data into predictive models for a wide range of biomedical applications, these models often fail even when the correct answer is already known, and fail to explain individual predictions in terms that data scientists can appreciate. These limitations suggest that new methods to produce practical artificial intelligence are still needed.
In this talk, I will discuss our work in (1) building an integrative knowledge infrastructure to prepare FAIR and "AI-ready" data and services along with (2) neurosymbolic AI methods to improve the quality of predictions and to generate plausible explanations. Attention is given to standards, platforms, and methods to wrangle knowledge into simple, but effective semantic and latent representations, and to make these available into standards-compliant and discoverable interfaces that can be used in model building, validation, and explanation. Our work, and those of others in the field, creates a baseline for building trustworthy and easy to deploy AI models in biomedicine.
Bio
Dr. Michel Dumontier is the Distinguished Professor of Data Science at Maastricht University, founder and executive director of the Institute of Data Science, and co-founder of the FAIR (Findable, Accessible, Interoperable and Reusable) data principles. His research explores socio-technological approaches for responsible discovery science, which includes collaborative multi-modal knowledge graphs, privacy-preserving distributed data mining, and AI methods for drug discovery and personalized medicine. His work is supported through the Dutch National Research Agenda, the Netherlands Organisation for Scientific Research, Horizon Europe, the European Open Science Cloud, the US National Institutes of Health, and a Marie-Curie Innovative Training Network. He is the editor-in-chief for the journal Data Science and is internationally recognized for his contributions in bioinformatics, biomedical informatics, and semantic technologies including ontologies and linked data.
Richard's entangled aventures in wonderlandRichard Gill
Since the loophole-free Bell experiments of 2020 and the Nobel prizes in physics of 2022, critics of Bell's work have retreated to the fortress of super-determinism. Now, super-determinism is a derogatory word - it just means "determinism". Palmer, Hance and Hossenfelder argue that quantum mechanics and determinism are not incompatible, using a sophisticated mathematical construction based on a subtle thinning of allowed states and measurements in quantum mechanics, such that what is left appears to make Bell's argument fail, without altering the empirical predictions of quantum mechanics. I think however that it is a smoke screen, and the slogan "lost in math" comes to my mind. I will discuss some other recent disproofs of Bell's theorem using the language of causality based on causal graphs. Causal thinking is also central to law and justice. I will mention surprising connections to my work on serial killer nurse cases, in particular the Dutch case of Lucia de Berk and the current UK case of Lucy Letby.
Introduction:
RNA interference (RNAi) or Post-Transcriptional Gene Silencing (PTGS) is an important biological process for modulating eukaryotic gene expression.
It is highly conserved process of posttranscriptional gene silencing by which double stranded RNA (dsRNA) causes sequence-specific degradation of mRNA sequences.
dsRNA-induced gene silencing (RNAi) is reported in a wide range of eukaryotes ranging from worms, insects, mammals and plants.
This process mediates resistance to both endogenous parasitic and exogenous pathogenic nucleic acids, and regulates the expression of protein-coding genes.
What are small ncRNAs?
micro RNA (miRNA)
short interfering RNA (siRNA)
Properties of small non-coding RNA:
Involved in silencing mRNA transcripts.
Called “small” because they are usually only about 21-24 nucleotides long.
Synthesized by first cutting up longer precursor sequences (like the 61nt one that Lee discovered).
Silence an mRNA by base pairing with some sequence on the mRNA.
Discovery of siRNA?
The first small RNA:
In 1993 Rosalind Lee (Victor Ambros lab) was studying a non- coding gene in C. elegans, lin-4, that was involved in silencing of another gene, lin-14, at the appropriate time in the
development of the worm C. elegans.
Two small transcripts of lin-4 (22nt and 61nt) were found to be complementary to a sequence in the 3' UTR of lin-14.
Because lin-4 encoded no protein, she deduced that it must be these transcripts that are causing the silencing by RNA-RNA interactions.
Types of RNAi ( non coding RNA)
MiRNA
Length (23-25 nt)
Trans acting
Binds with target MRNA in mismatch
Translation inhibition
Si RNA
Length 21 nt.
Cis acting
Bind with target Mrna in perfect complementary sequence
Piwi-RNA
Length ; 25 to 36 nt.
Expressed in Germ Cells
Regulates trnasposomes activity
MECHANISM OF RNAI:
First the double-stranded RNA teams up with a protein complex named Dicer, which cuts the long RNA into short pieces.
Then another protein complex called RISC (RNA-induced silencing complex) discards one of the two RNA strands.
The RISC-docked, single-stranded RNA then pairs with the homologous mRNA and destroys it.
THE RISC COMPLEX:
RISC is large(>500kD) RNA multi- protein Binding complex which triggers MRNA degradation in response to MRNA
Unwinding of double stranded Si RNA by ATP independent Helicase
Active component of RISC is Ago proteins( ENDONUCLEASE) which cleave target MRNA.
DICER: endonuclease (RNase Family III)
Argonaute: Central Component of the RNA-Induced Silencing Complex (RISC)
One strand of the dsRNA produced by Dicer is retained in the RISC complex in association with Argonaute
ARGONAUTE PROTEIN :
1.PAZ(PIWI/Argonaute/ Zwille)- Recognition of target MRNA
2.PIWI (p-element induced wimpy Testis)- breaks Phosphodiester bond of mRNA.)RNAse H activity.
MiRNA:
The Double-stranded RNAs are naturally produced in eukaryotic cells during development, and they have a key role in regulating gene expression .
Slide 1: Title Slide
Extrachromosomal Inheritance
Slide 2: Introduction to Extrachromosomal Inheritance
Definition: Extrachromosomal inheritance refers to the transmission of genetic material that is not found within the nucleus.
Key Components: Involves genes located in mitochondria, chloroplasts, and plasmids.
Slide 3: Mitochondrial Inheritance
Mitochondria: Organelles responsible for energy production.
Mitochondrial DNA (mtDNA): Circular DNA molecule found in mitochondria.
Inheritance Pattern: Maternally inherited, meaning it is passed from mothers to all their offspring.
Diseases: Examples include Leber’s hereditary optic neuropathy (LHON) and mitochondrial myopathy.
Slide 4: Chloroplast Inheritance
Chloroplasts: Organelles responsible for photosynthesis in plants.
Chloroplast DNA (cpDNA): Circular DNA molecule found in chloroplasts.
Inheritance Pattern: Often maternally inherited in most plants, but can vary in some species.
Examples: Variegation in plants, where leaf color patterns are determined by chloroplast DNA.
Slide 5: Plasmid Inheritance
Plasmids: Small, circular DNA molecules found in bacteria and some eukaryotes.
Features: Can carry antibiotic resistance genes and can be transferred between cells through processes like conjugation.
Significance: Important in biotechnology for gene cloning and genetic engineering.
Slide 6: Mechanisms of Extrachromosomal Inheritance
Non-Mendelian Patterns: Do not follow Mendel’s laws of inheritance.
Cytoplasmic Segregation: During cell division, organelles like mitochondria and chloroplasts are randomly distributed to daughter cells.
Heteroplasmy: Presence of more than one type of organellar genome within a cell, leading to variation in expression.
Slide 7: Examples of Extrachromosomal Inheritance
Four O’clock Plant (Mirabilis jalapa): Shows variegated leaves due to different cpDNA in leaf cells.
Petite Mutants in Yeast: Result from mutations in mitochondrial DNA affecting respiration.
Slide 8: Importance of Extrachromosomal Inheritance
Evolution: Provides insight into the evolution of eukaryotic cells.
Medicine: Understanding mitochondrial inheritance helps in diagnosing and treating mitochondrial diseases.
Agriculture: Chloroplast inheritance can be used in plant breeding and genetic modification.
Slide 9: Recent Research and Advances
Gene Editing: Techniques like CRISPR-Cas9 are being used to edit mitochondrial and chloroplast DNA.
Therapies: Development of mitochondrial replacement therapy (MRT) for preventing mitochondrial diseases.
Slide 10: Conclusion
Summary: Extrachromosomal inheritance involves the transmission of genetic material outside the nucleus and plays a crucial role in genetics, medicine, and biotechnology.
Future Directions: Continued research and technological advancements hold promise for new treatments and applications.
Slide 11: Questions and Discussion
Invite Audience: Open the floor for any questions or further discussion on the topic.
A brief information about the SCOP protein database used in bioinformatics.
The Structural Classification of Proteins (SCOP) database is a comprehensive and authoritative resource for the structural and evolutionary relationships of proteins. It provides a detailed and curated classification of protein structures, grouping them into families, superfamilies, and folds based on their structural and sequence similarities.
4. Apa itu Hukum
Gauss?
Hukum yang menentukan besarnya fluks listrik atau kuat
medan listrik yang memalui sebuah bidang
Fluks listrik di suatu daerah didefiniskan sebagai medan listrik
dikalikan dengan luas permukaan yang diproyeksikan dalam
bidang tegak lurus terhadap medan. Artinya jumlah garis gaya
yang digunakan untuk menembus suatu permukaan tertutup
adalah sama dengan muatan
Apabila terdapat garis suatu medan listrik yang homogen
menembus suatu bidang “A”, maka fluks listrik yang melewati
bidang tersebut bergantung pada kuatnya medan listrik, sudut
yang jatuh serta luas bidang yang ditembus
5. Electric Flux
We define the electric flux ,
of the electric field E,
through the surface A, as:
= E . A
Where:
A is a vector normal to the surface
(magnitude A, and direction normal to the surface).
is the angle between E and A
area A
E
A
= E A cos ()
6. Electric Flux
Here the flux is
= E · A
You can think of the flux through some surface as a measure of
the number of field lines which pass through that surface.
Flux depends on the strength of E, on the surface area, and on
the relative orientation of the field and surface.
Normal to surface,
magnitude A
area A
E
A
E
A
7. Electric Flux
The flux also depends on orientation
area A
A cos
The number of field lines through the tilted surface equals the
number through its projection . Hence, the flux through the tilted
surface is simply given by the flux through its projection: E (A cos).
= E . A = E A cos
area A
A cos
E E
A A
8. Exercise 1:
Calculate the flux of the electric field E,
through the surface A, in each of the
three cases shown:
a) =
b) =
c) =
9.
dA
E
What if the surface is curved, or the field varies with position ??
1. We divide the surface into small
regions with area dA
2. The flux through dA is
d = E dA cos
d = E . dA
3. To obtain the total flux we need
to integrate over the surface A
A
= d = E . dA
= E . A
10. In the case of a closed surface
d E dA
q
inside
0
The loop means the integral is over a closed surface.
dA
E
11. For a closed surface:
The flux is positive for field lines
that leave the enclosed volume
The flux is negative for field lines
that enter the enclosed volume
If a charge is outside a closed surface, the net flux is zero.
As many lines leave the surface, as lines enter it.
12. Exercise 2:
For which of these closed surfaces (a, b, c, d)
the flux of the electric field, produced by the
charge +2q, is zero?
13.
14. Spherical surface with point charge at center
d E dA
Flux of electric field:
2
0
1
cos
4
q dA
E dA
r
2
but , then:
dA
d
r
0 0 0
4
4 4
q q q
d
0
Gauss's Law
enclosed
q
E dA
15. Gauss’s Law
The electric flux
through any closed surface
equals enclosed charge / 0
"fluks total yang melewati permukaan tertutup sebanding
dengan muatan total yang dilingkupi permukaan tertutup
tersebut"
d E dA
q
inside
0
16. Exercise 3:
Calculate the flux of the electric field
for each of the closed surfaces a, b, c, and d
Surface a, a =
Surface b, b =
Surface c, c =
Surface d, d =
17. Calculate the electric field produced
by a point charge using Gauss Law
We choose for the gaussian surface a sphere
(bola) of radius r, centered on the charge Q.
Then, the electric field E, has the same
magnitude everywhere on the surface
(radial symmetry)
Furthermore, at each point on the surface,
the field E and the surface normal dA are
parallel (both point radially outward).
E . dA = E dA [cos = 1]
18. E
Q
Coulomb’s Law !
E
1
40
q
r 2
E . dA = Q / 0
E . dA = E dA = E A
A = 4 r2
E A = E 4 r2 = Q / 0
Electric field produced
by a point charge
E
Q
k = 1 / 4 0
0 = permittivity
0 = 8.85x10-12 C2/Nm2
19. E d
A =
Qenclosed
0
Gauss’s Law
The total flux within
a closed surface …
… is proportional to
the enclosed charge.
Gauss’s Law is always true, but is only useful for certain
very simple problems with great symmetry.
20. Applying Gauss’s Law
Gauss’s law is useful only when the electric field
is constant on a given surface
Infinite sheet of charge
1. Select Gauss surface
In this case a cylindrical
pillbox
2. Calculate the flux of the
electric field through the
Gauss surface
= 2 E A
3. Equate = qencl/0
2EA = qencl/0
4. Solve for E
E = qencl / 2 A 0 = / 2 0
(with = qencl / A)
21. GAUSS LAW – SPECIAL SYMMETRIES
SPHERICAL
(point or sphere)
CYLINDRICAL
(line or cylinder)
PLANAR
(plane or sheet)
CHARGE
DENSITY
Depends only on
radial distance
from central point
Depends only on
perpendicular distance
from line
Depends only on
perpendicular distance
from plane
GAUSSIAN
SURFACE
Sphere centered at
point of symmetry
Cylinder centered at
axis of symmetry
Pillbox or cylinder
with axis
perpendicular to plane
ELECTRIC
FIELD E
E constant at
surface
E ║A - cos = 1
E constant at curved
surface and E ║ A
E ┴ A at end surface
cos = 0
E constant at end
surfaces and E ║ A
E ┴ A at curved surface
cos = 0
FLUX
23. A charge Q is uniformly distributed through a sphere of radius R.
What is the electric field as a function of r?. Find E at r1 and r2.
Problem: Sphere of Charge Q
r2
r1
R
24. A charge Q is uniformly distributed through a sphere of radius R.
What is the electric field as a function of r?. Find E at r1 and r2.
Problem: Sphere of Charge Q
Use symmetry!
This is spherically symmetric.
That means that E(r) is radially
outward, and that all points, at a
given radius (|r|=r), have the same
magnitude of field.
r2
r1
R
E(r1)
E(r2)
25. Problem: Sphere of Charge Q
E & dA
r
R
What is the enclosed charge?
First find E(r) at a point outside the charged sphere. Apply Gauss’s
law, using as the Gaussian surface the sphere of radius r pictured.
26. Problem: Sphere of Charge Q
r
R
E & dA What is the enclosed charge? Q
First find E(r) at a point outside the charged sphere. Apply Gauss’s
law, using as the Gaussian surface the sphere of radius r pictured.
27. Problem: Sphere of Charge Q
r
R
E & dA What is the enclosed charge? Q
What is the flux through this surface?
First find E(r) at a point outside the charged sphere. Apply Gauss’s
law, using as the Gaussian surface the sphere of radius r pictured.
28. Problem: Sphere of Charge Q
r
R
E & dA What is the enclosed charge? Q
What is the flux through this surface?
E d
A EdA
E dA
EA E(4 r2
)
First find E(r) at a point outside the charged sphere. Apply Gauss’s
law, using as the Gaussian surface the sphere of radius r pictured.
29. Problem: Sphere of Charge Q
r
R
E & dA What is the enclosed charge? Q
What is the flux through this surface?
E d
A EdA
E dA
EA E(4 r2
)
Gauss Q /o
Q/0 E(4 r
2
)
First find E(r) at a point outside the charged sphere. Apply Gauss’s
law, using as the Gaussian surface the sphere of radius r pictured.
30. Problem: Sphere of Charge Q
r
R
E & dA What is the enclosed charge? Q
What is the flux through this surface?
E d
A EdA
E dA
EA E(4 r2
)
Gauss: Q /o
So E(r)
1
4o
Q
r
2
ˆ
r
Exactly as though all the
charge were at the origin!
(for r>R)
Q/0 E(4 r
2
)
First find E(r) at a point outside the charged sphere. Apply Gauss’s
law, using as the Gaussian surface the sphere of radius r pictured.
31. Problem: Sphere of Charge Q
R
r
E(r
)
Next find E(r) at a point inside the sphere. Apply Gauss’s law,
using a little sphere of radius r as a Gaussian surface.
What is the enclosed charge?
That takes a little effort. The little sphere has
some fraction of the total charge. What fraction?
That’s given by volume ratio: Qenc
r3
R
3 Q
Again the flux is: = EA = E(4 r 2
)
Setting Qenc /o gives E =
(r3
/R3
)Q
4or
2
E(r) =
Q
4oR3 r ˆ
r
For r<R
33. Problem: Sphere of Charge Q
Look closer at these results. The electric field at comes
from a sum over the contributions of all the little bits .
It’s obvious that the net E at this point will be horizontal.
But the magnitude from each bit is different; and it’s completely
not obvious that the magnitude E just depends on the distance
from the sphere’s center to the observation point.
Doing this as a volume integral would be HARD.
Gauss’s law is EASY.
R
Q r
r > R
34.
Problem: Infinite charged plane
Consider an infinite plane with a constant surface charge density
(which is some number of Coulombs per square meter).
What is E at a point located a distance z above the plane?
x
y
z
35.
Problem: Infinite charged plane
Consider an infinite plane with a constant surface charge density
(which is some number of Coulombs per square meter).
What is E at a point located a distance z above the plane?
x
y
z
Use symmetry!
The electric field must point straight away
from the plane (if > 0). Maybe the
Magnitude of E depends on z, but the direction
is fixed. And E is independent of x and y.
E
36. E
E
Gaussian “pillbox”
Problem: Infinite charged plane
So choose a Gaussian surface that is a “pillbox”, which has its top
above the plane, and its bottom below the plane, each a distance z
from the plane. That way the observation point lies in the top.
z
z
43. E
E
Gaussian “pillbox”
Problem: Infinite charged plane
z
z
Let the area of the top and bottom be A.
Outward flux through the top: EA
Outward flux through the bottom: EA
44. E
E
Gaussian “pillbox”
Problem: Infinite charged plane
z
z
Let the area of the top and bottom be A.
Outward flux through the top: EA
Outward flux through the bottom: EA
Outward flux through the sides:
45. E
E
Gaussian “pillbox”
Problem: Infinite charged plane
z
z
Let the area of the top and bottom be A.
Outward flux through the top: EA
Outward flux through the bottom: EA
Outward flux through the sides: E x (some area) x cos(900) = 0
46. E
E
Gaussian “pillbox”
Problem: Infinite charged plane
z
z
Outward flux through the top: EA
Outward flux through the bottom: EA
Outward flux through the sides: E x (some area) x cos(900) = 0
So the total flux is: 2EA
Let the area of the top and bottom be A.
47. E
E
Gaussian “pillbox”
Problem: Infinite charged plane
z
z
Gauss’s law then says that A/0=2EA so that E=/20, outward.
This is constant everywhere in each half-space!
Let the area of the top and bottom be A.
Notice that the area A canceled: this is typical!
48. Problem: Infinite charged plane
Imagine doing this with an integral over the charge distribution:
break the surface into little bits dA …
dE
Doing this as a surface integral would be HARD.
Gauss’s law is EASY.
49. Consider a long cylindrical charge distribution of radius R,
with charge density = a – b r (with a and b positive).
Calculate the electric field for:
a) r < R
b) r = R
c) r > R
50. • A conductor is a material in which charges can move
relatively freely.
• Usually these are metals (Au, Cu, Ag, Al).
• Excess charges (of the same sign) placed on a conductor
will move as far from each other as possible, since they
repel each other.
• For a charged conductor, in a static situation, all the
charge resides at the surface of a conductor.
• For a charged conductor, in a static situation, the electric
field is zero everywhere inside a conductor, and
perpendicular to the surface just outside
Conductors
52. Conductors
Why is E = 0 inside a conductor?
Conductors are full of free electrons, roughly one per
cubic Angstrom. These are free to move. If E is
nonzero in some region, then the electrons there feel
a force -eE and start to move.
53. Conductors
Why is E = 0 inside a conductor?
Conductors are full of free electrons, roughly one per
cubic Angstrom. These are free to move. If E is
nonzero in some region, then the electrons there feel
a force -eE and start to move.
In an electrostatics problem, the electrons adjust their
positions until the force on every electron is zero (or
else it would move!). That means when equilibrium is
reached, E=0 everywhere inside a conductor.
54. Conductors
Because E = 0 inside, the inside of a conductor is neutral.
Suppose there is an extra charge inside.
Gauss’s law for the little spherical surface
says there would be a nonzero E nearby.
But there can’t be, within a metal!
Consequently the interior of a metal is neutral.
Any excess charge ends up on the surface.
55. Electric field just outside a charged conductor
0 0
0
enclosed
E dA EA
q A
EA
E
The electric field just outside a charged conductor
is perpendicular to the surface and has magnitude E = / 0
56. Properties of Conductors
In a conductor there are large number of electrons free to move.
This fact has several interesting consequences
Excess charge placed on a conductor moves to the exterior
surface of the conductor
The electric field inside a conductor is zero when charges
are at rest
A conductor shields a cavity within it from external electric
fields
Electric field lines contact conductor surfaces at right angles
A conductor can be charged by contact or induction
Connecting a conductor to ground is referred to as grounding
The ground can accept of give up an unlimited number of electrons
57.
58. a
b
Problem: Charged coaxial cable
This picture is a cross section of an infinitely long line of charge,
surrounded by an infinitely long cylindrical conductor. Find E.
This represents the line of charge.
Say it has a linear charge density of l
(some number of C/m).
This is the cylindrical conductor. It
has inner radius a, and outer radius b.
Use symmetry!
Clearly E points straight out, and its
amplitude depends only on r.
59. Problem: Charged coaxial cable
First find E at positions in the space inside the cylinder (r<a).
Choose as a Gaussian surface:
a cylinder of radius r, and length L.
r
L
60. What is the charge enclosed? lL
What is the flux through the end caps? zero (cos900)
What is the flux through the curved face? E x (area) = E(2rL)
Total flux = E(2rL)
Gauss’s law E(2rL) = lL/0 so E(r) = l/ 2r0
Problem: Charged coaxial cable
First find E at positions in the space inside the cylinder (r<a).
r
L
61. Problem: Charged coaxial cable
Now find E at positions within the cylinder (a<r<b).
Make the same kind of cylindrical Gaussian
surface. Now the curved side is entirely
within the conductor, where E=0; hence the
flux is zero.
There’s no work to do: within a conductor E=0.
Still, we can learn something from Gauss’s law.
r
+
Thus the total charge
enclosed by this surface
must be zero.
62. Problem: Charged coaxial cable
There must be a net charge per unit length -l
attracted to the inner surface of the metal, so
that the total charge enclosed by this Gaussian
surface is zero.
r
+
-
-
-
-
-
-
63. Problem: Charged coaxial cable
+
+
r
+
-
-
-
-
-
-
+
+
+
+
There must be a net charge per unit length
–l attracted to the inner surface of the
metal so that the total charge enclosed by
this Gaussian surface is zero.
And since the cylinder is neutral, these
negative charges must have come from
the outer surface. So the outer surface
has a charge density per unit length of
+l spread around the outer perimeter.
So what is the field for r>b?