Electric field of an infinite charged plate

  • Thread starter jisbon
  • Start date
  • #26
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,851
6,120
Not sure what you meant, but won't it repel away from its point? for eg:

View attachment 249633
It is an infinite plate. Why would the repulsion be down (in terms of your diagram) rather than up?
 
  • #27
464
30
Not sure what you meant, but won't it repel away from its point? for eg:

View attachment 249633
1568441652637.png

Pardon for messy drawing, but if I'm correct, the charge particle simply repels to the right right? Since the vertical components are pretty much cancelled out
 
  • Like
Likes Delta2 and PeroK
  • #28
464
30
Seems like my idea was correct. However, in my initial attempt, I have already understood that vertical forces will be canceled. So hence my resultant force:
force from plate (only affects x axis) + force from particle (only affects y axis)?
 
  • #29
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,851
6,120
Seems like my idea was correct. However, in my initial attempt, I have already understood that vertical forces will be canceled. So hence my resultant force:
force from plate (only affects x axis) + force from particle (only affects y axis)?
Yes, but in post #14 you had the plate generating a field in the z axis also.
 
  • #30
464
30
Yes, but in post #14 you had the plate generating a field in the z axis also.
Oh.. my bad. So its:

##E_{(4,4,0)}=E_{plate} + E_{particle} = (4519.77\widetilde{x})N/C +(-1125\widetilde{y})N/C##?
 
  • #31
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,851
6,120
Oh.. my bad. So its:

##E_{(4,4,0)}=E_{plate} + E_{particle} = (4519.77\widetilde{x})N/C +(-1125\widetilde{y})N/C##?
Yes.
 
  • #32
464
30
Yes.
I've calculated the magnitude and it appears to be 4657, which is different from the given answer, which is 9109 :/
 
  • #33
TSny
Homework Helper
Gold Member
12,852
3,203
I think your answer is correct if the infinite plate has only one face with charge density ##\sigma##. However, if they meant the plate has two faces, each face having charge density 8 x 10-8 C/m2, then you would get their answer. If this latter interpretation is what they meant, then I misled you with the diagram in post #2. Sorry about that.
 
Last edited:
  • #34
464
30
I think your answer is correct if the infinite plate has only one face with charge density ##\sigma##. However, if they meant the plate has two faces, each face having charge density 8 x 10-8 C/m2, then you would get their answer. If this latter interpretation is what they meant, then I misled you with the diagram in post #2. Sorry about that.
Hmm okay. So if it has two faces, is there now supposed to be a negative x direction now? Not sure how to proceed
 
  • #35
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,851
6,120
Hmm okay. So if it has two faces, is there now supposed to be a negative x direction now? Not sure how to proceed
If there is a given charge density on an infinite plate then the flux lines go equally from both sides. So the field strength is half what it would be if all the flux lines emerged from the same side. This leads to the factor ##\frac 12## in the formula.
But in this problem there is an ambiguity. It gives the "surface" charge density. Since a plate has two surfaces, they might mean that each surface has that density.
 
  • #36
464
30
If there is a given charge density on an infinite plate then the flux lines go equally from both sides. So the field strength is half what it would be if all the flux lines emerged from the same side. This leads to the factor ##\frac 12## in the formula.
But in this problem there is an ambiguity. It gives the "surface" charge density. Since a plate has two surfaces, they might mean that each surface has that density.
So if I'm following what you are saying, won't dividing the field strength by plate by 2 make the magnitude even smaller?

In which: ##E_{(4,4,0)}=E_{plate} + E_{particle} = (4519.77/2 \widetilde{x})N/C +(-1125\widetilde{y})N/C##

Even if it's the other scenario (whereby I multiply by 2), the magnitude of the Efield is still 9107.7N/C , which is due to some form of rounding up/down I presume?
 
  • #37
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,851
6,120
won't dividing the field strength by plate by 2 make the magnitude even smaller?
No, you already did that divide by 2 in your post #3 to arrive at 4519... Look at the equation you used there.
I am saying that if the question means the given charge density exists on both sides of the plate then there is in effect twice the charge density, so the 1/2 should not be in the.formula.
 
  • #38
464
30
No, you already did that divide by 2 in your post #3 to arrive at 4519... Look at the equation you used there.
I am saying that if the question means the given charge density exists on both sides of the plate then there is in effect twice the charge density, so the 1/2 should not be in the.formula.
Oh ok.

But even if I don't divide it by 2,

Even if it's the other scenario (whereby I multiply by 2), the magnitude of the Efield is still 9107.7N/C , which is due to some form of rounding up/down I presume?
Maybe an rounding off error or?
 
  • #39
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,851
6,120
Oh ok.

But even if I don't divide it by 2,



Maybe an rounding off error or?
Using your numbers for the two components I get 9209.
What is the supposed answer?
 
  • #40
464
30
Using your numbers for the two components I get 9209.
What is the supposed answer?
9109
 
  • #42
TSny
Homework Helper
Gold Member
12,852
3,203
I happened to notice that you get 9109 N/C for E if you use ##\epsilon_0 = 8.85 \times 10^{-12}## and ##k = \frac{1}{4 \pi \epsilon_0}##. But this is kind of silly as it rounds ##\epsilon_0## to 3 significant figures in order to try to get an answer accurate to 4 sig figs! And the data is given to only 1 sig fig.
 
  • #43
464
30
I happened to notice that you get 9109 N/C for E if you use ##\epsilon_0 = 8.85 \times 10^{-12}## and ##k = \frac{1}{4 \pi \epsilon_0}##. But this is kind of silly as it rounds ##\epsilon_0## to 3 significant figures in order to try to get an answer accurate to 4 sig figs! And the data is given to only 1 sig fig.
Possibly a typo.
Ah alright. Guess it's probably an accuracy error then.
 

Related Threads on Electric field of an infinite charged plate

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
1K
Replies
3
Views
1K
Replies
4
Views
1K
Replies
1
Views
10K
Replies
11
Views
240
Replies
8
Views
11K
Replies
6
Views
1K
Replies
4
Views
34K
Top