- #36

jisbon

- 476

- 30

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?