# Electric field of an infinite charged plate

• jisbon
In summary: If you're saying that at the point (4, 4, 0) the electric field due to the point charge is parallel to the y-axis and the field due to the infinite plane is parallel to the x-axis, that's correct. So, how do you add two fields that point in different... directions?
haruspex said:
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?

jisbon said:
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.

haruspex said:
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,

jisbon said:
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?

jisbon said:
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.

haruspex said:
Using your numbers for the two components I get 9209.
9109

jisbon said:
9109
Possibly a typo.

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.

TSny said:
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.
haruspex said:
Possibly a typo.
Ah alright. Guess it's probably an accuracy error then.

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