Matching Electric Field Entries

In summary: The field was not zero in part b). But in the present situation, there are two spheres, a charged point particle and a metal shell with inner void. The charge q1 induces some charge on the outer surface of the shell, but that outer field does not penetrate inside the metal wall. Inside, the field is due to the point charge only and it is zero. If there is no charge inside the metal wall, the field is zero. ehildThe field was not zero in part b). But in the present situation, there are two spheres, a charged point particle and a metal shell with inner void. The charge q1 induces some charge on the outer surface of the shell, but that outer field does not penetrate inside the metal
  • #1
Saitama
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Homework Statement


Match the entries of Column-I with entries of Column-II
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E is the electric field vector.
There can be more than one answer for each entry in column-I.

Homework Equations


The Attempt at a Solution


I am not sure about any of the answer and i don't have the answer key too.
I would like to discuss each entry in Column-I one by one. :smile:
For the first question, i suppose the answer should be r) and t) because the conductor is neutral and the potential will be due to the point charge and electric field will be produced by the point charge kept near to the conductor. Is my reasoning correct?

EDIT: 800th post. :D
 
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  • #2
Congratulation for the 800th post. But why do you think that the magnitude of E varies in the void inside the conductor if there is no charge there? The charge is outside. What does Gauss' law state? ehild
 
  • #3
Pranav-Arora said:
the conductor is neutral and the potential will be due to the point charge and electric field will be produced by the point charge kept near to the conductor. Is my reasoning correct?
there will also be the contribution of induced charges on the conductor surface (the net induced charge being zero) on the potential and field inside it.
the field inside a conductor is independent of the charges outside it.if you bring another charge outside ,although the distribution of charge on conductor surface will change but the field inside will remain same.
in fact in this case field is zero.
 
  • #4
ehild said:
Congratulation for the 800th post. But why do you think that the magnitude of E varies in the void inside the conductor if there is no charge there? The charge is outside. What does Gauss' law state?


ehild

Thank you ehild, i understand it now. By Gauss law there would be no electric field inside the conductor. But what about the potential? I think it should be constant because inside the conductor, E is zero, therefore -dV/dr=0 or dV=0, therefore V is constant. So the answer should be p) and s) ?
 
  • #5
If the electric field is zero the potential is constant. Moreover, the potential is a continuous function, it is the same at both sides of the inner surface of the metal shell, inside the void and inside the metal layer. You are right, p) and s) in column II matches to case a) in column I.
And what about q)?

ehild
 
  • #6
ehild said:
And what about q)?

Not sure about q), i guess |E| should also be zero. :confused:
 
  • #7
yes,|E| is also zero.
 
  • #8
pcm said:
yes,|E| is also zero.

Thanks! :smile:
Can i have some hints for part b)?
 
  • #9
Gauss? You surround the point charge with a sphere. Can E = 0 everywhere on its surface?

ehild
 
  • #10
ehild said:
Gauss? You surround the point charge with a sphere. Can E = 0 everywhere on its surface?

ehild

No, E is not zero on the surface if we surround the point charge with a sphere.
 
  • #11
Yes. So are p) and q) true if E somewhere inside the shell differs from zero?

What about r)?

ehild
 
  • #12
ehild said:
Yes. So are p) and q) true if E somewhere inside the shell differs from zero?

What about r)?

ehild

No p) and q) are not true.

r) should be the correct answer. As E is varying, potential should also vary, therefore the answer should be r) and t). Am i right?
 
  • #13
Yes. they are true, but can you explain why is |E| varying?

ehild
 
  • #14
ehild said:
Yes. they are true, but can you explain why is |E| varying?

ehild

Electric field inside the conductor is due to the point charge and it varies inversely with the square of distance from the charge, therefore the magnitude of E should also vary. Is my explanation correct?
 
  • #15
Almost.:smile:The point charge is not alone, it is surrounded by the metal shell and there is some induced surface charge distribution in the inner surface. The electric field is due to the point charge and that surface charge distribution. In general, the field is not the same as that of a single point charge, but because of spherical symmetry, it is.

There comes the next step, c. What do you think?

ehild
 
  • #16
ehild said:
Almost.:smile:The point charge is not alone, it is surrounded by the metal shell and there is some induced surface charge distribution in the inner surface. The electric field is due to the point charge and that surface charge distribution. In general, the field is not the same as that of a single point charge, but because of spherical symmetry, it is.

There comes the next step, c. What do you think?

ehild

Thanks ehild! :smile:

About c, i am confused, it seems like the combination of both parts, a) and b). Applying Gauss law and surrounding the charge q2 with a sphere, E is not zero on the surface of sphere. E is varying inside and hence the potential is also varying. The answer should be again r) and t) but i think i am doing something wrong as their is still a charge q1 present. q1 is scaring me.
 
  • #17
Do not be scared by q1. It induces a charge distribution on the outer surface of the metal shell, but the field is still zero inside the metal wall. The outer field does not penetrate into the void. The point charge q2 inside does not feel anything, it just does not know about the presence of q1 so it is not scared of it, why are you?

ehild
 
  • #18
ehild said:
Do not be scared by q1. It induces a charge distribution on the outer surface of the metal shell, but the field is still zero inside the metal wall. The outer field does not penetrate into the void. The point charge q2 inside does not feel anything, it just does not know about the presence of q1 so it is not scared of it, why are you?

ehild

How the electric field is zero? :confused:
In the part b), E was present in conductor when a charge is kept inside but why here it is zero? I don't get the point here. :(
 
  • #19
The electric field is zero in the metal wall, not in the void it surrounds.

ehild
 
  • #20
ehild said:
The electric field is zero in the metal wall, not in the void it surrounds.

ehild

What about the void then? :smile:
 
  • #21
There is a point charge in the middle... And it does not know anything about the outer world as it is shielded from it by the metal shell.

ehild
 
  • #22
ehild said:
There is a point charge in the middle... And it does not know anything about the outer world as it is shielded from it by the metal shell.

ehild

The answer should be again r) and t) because if the charge q2 knows nothing about the outer world, the case becomes same as part b). Am i right?
 
  • #23
It looks so... ehild
 
  • #24
ehild said:
It looks so...


ehild

Looks so? Did i say something wrong?
 
  • #25
No, but what if I am not right? I can not say that "yes, it is the right solution", because I can not be sure in myself (getting old, getting mad, remembering wrong) and you can not trust in me. But you can trust in your own knowledge and logic.:biggrin:

ehild
 
  • #26
ehild said:
...(getting old, getting mad, remembering wrong) and you can not trust in me. But you can trust in your own knowledge and logic.:biggrin:

ehild

Sorry ehild, if i annoyed you. :uhh:

I am not an intelligent student that i could always trust my logic and when it comes to physics, i am handicapped. I always need to double check the stuff i do. :smile:

Can we move on to the next question and i will again say the answer to part c) are r) and t).
 
  • #27
I would say the same. If it is enough to you...
But what would you answer if your teacher asks why? Would you say that it is so because an unknown person from a far-away little country said so?
Have you learned something about a Faraday cage? You can mention that.

ehild
 
  • #28
ehild said:
I would say the same. If it is enough to you...
But what would you answer if your teacher asks why? Would you say that it is so because an unknown person from a far-away little country said so?
:rofl: :rofl:
Little country? I think Hungary is big enough. :smile:

ehild said:
Have you learned something about a Faraday cage? You can mention that.
No, i don't think so. There's nothing like Faraday cage mentioned in my book till the chapters i have read.
 
  • #30
ehild said:

Thank you ehild for that link, i read that page, seems like it would be helpful in solving the d part. ^_^

The conductor is charged this time. The field inside the conductor is varying due to the point charge, the potential is also varying, the same story repeats. q2 doesn't know about the outside world and the answers are again r) and t). Am i right? :biggrin:
 
  • #31
You are right. And you seem to like my explanation based on a personified point charge who does not know about the outside word.:biggrin: (Because it is locked into a Faraday cage, the poor thing.)

ehild
 
  • #32
ehild said:
You are right. And you seem to like my explanation based on a personified point charge who does not know about the outside world.:biggrin: (Because it is locked into a Faraday cage, the poor thing.)

ehild

Oh yes, i like that explanation. :smile:

Thank you ehild for all the help, the answers are really interesting. Three parts share the same answers.
 

1. What is an electric field?

An electric field is a physical field that surrounds a charged particle and exerts a force on other charged particles within the field. It is a fundamental concept in electromagnetism and is a key factor in understanding the behavior of electrically charged particles.

2. How is an electric field represented?

An electric field is represented by a vector, with both magnitude and direction. The magnitude of the electric field is determined by the strength of the charge that creates the field, while the direction of the field is determined by the direction the force would act on a positively charged particle placed in the field.

3. What is the unit of measurement for electric field?

The unit of measurement for electric field is newtons per coulomb (N/C) in SI units. This unit represents the force exerted on a unit of charge within the field.

4. How are electric field lines used to visualize electric fields?

Electric field lines are used to visualize electric fields by showing the direction and relative strength of the field at different points. The lines are drawn such that they are tangent to the direction of the electric field at each point, and the density of the lines represents the strength of the field.

5. What is the significance of matching electric field entries?

Matching electric field entries is important for understanding and predicting the behavior of charged particles within a given electric field. By matching the magnitude and direction of the electric field at different points, we can determine the force that would act on a charged particle placed at that point, and thus understand the motion and interactions of charged particles in the field.

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