Electric field of an infinite charged plate

[haruspex]

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.

 

[jisbon]

Yes.

I've calculated the magnitude and it appears to be 4657, which is different from the given answer, which is 9109 :/

 

[TSny]

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.

 

[jisbon]

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

 

[haruspex]

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.

 

[jisbon]

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?

 

[haruspex]

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.

 

[jisbon]

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?

 

[haruspex]

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?

 

[jisbon]

Using your numbers for the two components I get 9209.
What is the supposed answer?

9109

 

[haruspex]

9109

Possibly a typo.

 

[TSny]

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.

 

[jisbon]

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.