Induction confusion for induced charge inside a metal conductor

  • #1
tellmesomething
324
38
Homework Statement
Induced charge inside a metal conductor infinitely bounded due to a positive charge q kept beside it is -q(1-1/k).
Relevant Equations
None
Following the above statement my teacher was trying to prove something and it started with suppose q1 and q2 are charges placed in a medium (k) of infinite expanse. The distance between them being r. He took q1 and q2 to be some spherical particles and not point charges and concluded that

Net force on q1= Force due to q2 + force due to Induced charges in q2 + force due to induced charges in q1.

Firstly I dont understand why Induced charges will have a different force (?) I dont even understand induction mathematically I think.
Secondly If I pretend that I understand it , the force due to Induced charges in q2 is said to be -q2 (1-1/k) but shouldn't it be -q1(1-1/k)??? Since the other charge is responsible for creating the induction.....
 
Physics news on Phys.org
  • #2
tellmesomething said:
Net force on q1= Force due to q2 + force due to Induced charges in q2 + force due to induced charges in q1.
I would guess that should read "Net force on q1= Force due to q2 + force due to Induced charges in the medium by q2 + force due to induced charges in the medium by q1."
Which all seems reasonable.
 
  • Like
Likes tellmesomething
  • #3
haruspex said:
I would guess that should read "Net force on q1= Force due to q2 + force due to Induced charges in the medium by q2 + force due to induced charges in the medium by q1."
Which all seems reasonable.
I see that does make some sense.
Follow up question, so induced charges in the medium by q1 was spread out in a sphere and so shouldnt net force due to this on q1 becomes 0 because its uniformly spread..... Furthermore then from the rest of the expression we can prove how K×epsilon (permittivity in vaccum) comes into picture when talking about net electrostatic force in a medium.
 
  • #4
haruspex said:
I would guess that should read "Net force on q1= Force due to q2 + force due to Induced charges in the medium by q2 + force due to induced charges in the medium by q1."
Which all seems reasonable.
What is still a bit of a mystery to me is how we get -q(1-1/k) as the induced charge expression...is it experimentally derived.....? Also what about the positive charges that are left out, they would also effect the net force right? Maybe because they might be at a very large distance in a medium of infinite expanses so we ignore it... Neither my book nor my teacher included the force due to induced positive charge?
 
  • #5
I don’t understand what k is. It makes the medium sound more like a dielectric than a conductor.
 
  • #6
haruspex said:
I don’t understand what k is. It makes the medium sound more like a dielectric than a conductor.
Yes, I think that is what it is. If the medium becomes vacuum by setting k =1, this force goes to zero. However, I have trouble picturing the geometry. OP's statement "my teacher tried to prove something" is not descriptive of what's going on.
 
  • #7
kuruman said:
Yes, I think that is what it is. If the medium becomes vacuum by setting k =1, this force goes to zero. However, I have trouble picturing the geometry. OP's statement "my teacher tried to prove something" is not descriptive of what's going on.
Sorry. The proof in question is that net force on a charge due to another charge in a medium is q1q2/4πK£r²...
 
  • #8
haruspex said:
I don’t understand what k is. It makes the medium sound more like a dielectric than a conductor.
It is the dielectric constant though...
 
  • #9
tellmesomething said:
Sorry. The proof in question is that net force on a charge due to another charge in a medium is q1q2/4πK£r²...
Then you need to find the electric field generated by one charge, say ##q_1## using Gauss's law $$\int_S \mathbf {E}\cdot \mathbf {\hat n}~dA=\frac{q_{1}}{\epsilon}.$$ Once you know the electric field, the force on ##q_2## due to ##q_1## is easy to find.
 
  • #10
kuruman said:
Then you need to find the electric field generated by one charge, say ##q_1## using Gauss's law $$\int_S \mathbf {E}\cdot \mathbf {\hat n}~dA=\frac{q_{1}}{\epsilon}.$$ Once you know the electric field, the force on ##q_2## due to ##q_1## is easy to find.
I think the situation is more complicated than that.
For a dielectric plane lamina, the field is diminished within the lamina but there is no external consequence. I believe we can think of that situation as equivalent to an induced charge on each surface. Since those charges are equal and opposite, the induced fields cancel outside.
But here we have an infinite 3D extent, so the "far side" disappears to infinity, and the induced charge density there tends to zero. So is there a nonzero induced field within the cavity containing the other point charge?
 
  • #11
Is there a cavity? I interpreted post #7
tellmesomething said:
Sorry. The proof in question is that net force on a charge due to another charge in a medium is q1q2/4πK£r²...
to mean that the charges are embedded in a linear dielectric of permittivity £ dielectric constant K . The given "show that" expression q1q2/4πK£r² matches this interpretation if £ ≡ ε0.
 
Last edited:
  • #12
kuruman said:
Is there a cavity? I interpreted post #7

to mean that the charges are embedded in a linear dielectric of permittivity £ dielectric constant K . The given "show that" expression q1q2/4πK£r² matches this interpretation if £ ≡ ε0.
That was in post #7, but I don’t get how this relates to post #1.
 
  • #13
haruspex said:
That was in post #7, but I don’t get how this relates to post #1.
I don't get it either. That is why I asked OP to clarify in post #6
kuruman said:
I have trouble picturing the geometry. OP's statement "my teacher tried to prove something" is not descriptive of what's going on.
whereupon OP posted #7.
 
  • #14
kuruman said:
Is there a cavity? I interpreted post #7

to mean that the charges are embedded in a linear dielectric of permittivity £ dielectric constant K . The given "show that" expression q1q2/4πK£r² matches this interpretation if £ ≡ ε0.
Yes £ = epsilon knot. Sorry I could net reply earlier, i was confused.I have just started electrostatics so I dont know things like gauss law mentioned above yet
 
  • #15
tellmesomething said:
Yes £ = epsilon knot. Sorry I could net reply earlier, i was confused.I have just started electrostatics so I dont know things like gauss law mentioned above yet
Please clarify the relationship between posts #1 and #7. What happened to 1-1/k and induction? Should we ignore post #1?
 
  • #16
haruspex said:
Please clarify the relationship between posts #1 and #7. What happened to 1-1/k and induction? Should we ignore post #1?
No. Post 1 is how the derivation of the proofs (mentioned in post 7) starts. Why would you ignore that? Do You mean I can only understand 1-1/k after understanding gauss law?
 
  • #17
tellmesomething said:
Why would you ignore that?
Because until post #16 the relationship between posts #1 and #7 was not apparent.

How to prove the formula in post #7 depends what your starting point is, i.e. how k is defined. I don't mean merely that it is the dielectric constant; what does it mean in terms something physically observable?
E.g. if you were to define it in terms of a local reduction of the externally applied field then the equation in post #7 would follow immediately. So how is your teacher defining it? Perhaps in terms of the induced polarisation?
 
  • #18
haruspex said:
Because until post #16 the relationship between posts #1 and #7 was not apparent.

How to prove the formula in post #7 depends what your starting point is, i.e. how k is defined. I don't mean merely that it is the dielectric constant; what does it mean in terms something physically observable?
E.g. if you were to define it in terms of a local reduction of the externally applied field then the equation in post #7 would follow immediately. So how is your teacher defining it? Perhaps in terms of the induced polarisation?
I dont know that. I only understand that k = Permittivity in medium/ Permittivity in vaccum....
 
  • #19
haruspex said:
Because until post #16 the relationship between posts #1 and #7 was not apparent.

How to prove the formula in post #7 depends what your starting point is, i.e. how k is defined. I don't mean merely that it is the dielectric constant; what does it mean in terms something physically observable?
E.g. if you were to define it in terms of a local reduction of the externally applied field then the equation in post #7 would follow immediately. So how is your teacher defining it? Perhaps in terms of the induced polarisation?
Should I attach the whole proof?
 
  • #20
tellmesomething said:
Should I attach the whole proof?
That would be most helpful.
 
  • #21
haruspex said:
That would be most helpful.
Note (In a medium)
Suppose q1 and q2 are placed in a medium(k) of infinite expanse
IMG_20240502_101448.jpg

Net force on q1, ## \vec F ## = Force due to q2+ Force due to induced negative charge around q1+ Force due to induced negative charge around q2

Force due to induced negative charge around q1 =0 as its uniformly distributed


$$ \vec F= \frac {q1×q2} {4πE_0 r²} + \frac {q1 [ -q2[1-\frac{1} {k}]} {4πE_0r²} $$

$$ \frac {q1 q2 } { 4πkE_0r²}$$
 
  • #22
haruspex said:
That would be most helpful.
Sorry for the late reply I was figuring out how to type in latex
 
  • Like
Likes berkeman
  • #23
tellmesomething said:
Note (In a medium)
Suppose q1 and q2 are placed in a medium(k) of infinite expanseView attachment 344358
Net force on q1, ## \vec F ## = Force due to q2+ Force due to induced negative charge around q1+ Force due to induced negative charge around q2

Force due to induced negative charge around q1 =0 as its uniformly distributed


$$ \vec F= \frac {q1×q2} {4πE_0 r²} + \frac {q1 [ -q2[1-\frac{1} {k}]} {4πE_0r²} $$

$$ \frac {q1 q2 } { 4πkE_0r²}$$
Looks to me to be the same logic as in https://www.physicsforums.com/threads/derivation-of-induced-charge-on-a-dielectric.846504, but run in the opposite direction.
 

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
217
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
2
Replies
36
Views
595
  • Introductory Physics Homework Help
Replies
23
Views
859
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
21
Views
831
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
6K
  • Introductory Physics Homework Help
Replies
5
Views
12K
Back
Top