Let's say we take an infinite charge sheet in y-z with charge density $σ$ and a negatively charged particle $(–Q)$ in x axis. Then we know electric field due to the sheet is $$E= \frac{\sigma}{2\epsilon_0}$$ and it will be in pushing charged particle away from the sheet but here we are having a negatively charged particle, it will also be experiencing some attraction to the sheet which ultimately be altering the electric field or force value. but if I want to calculate the attraction force on this particle then can I just use coloumb's law or I have to follow any other method
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$\begingroup$ Is this infinitely charged sheet conducting or non-conducting? How much work have you put into this? This is very much a homework problem. $\endgroup$– naturallyInconsistentCommented May 7, 2023 at 12:25
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Assuming the sheet is nonconducting (so the formula $E = \frac{\sigma}{2\epsilon_0}$ is valid), and the charge density $\sigma$ is positive, the electric field is uniform and points away from the infinite sheet of charge. The force exerted on a negatively charged particle by the electric field is always in the opposite direction opposite to the electric field itself, so the particle will be attracted toward the sheet of charge. The force on the particle can be calculated using $\vec{F} = \vec{E}q$.
$$\vec{F} = \vec{E}(-Q) = -Q\frac{\sigma}{2\epsilon_0}$$
If the infinite sheet of charge was conducting then the distribution of charges would become more complicated.
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$\begingroup$ This cannot be correct. If the sheet is conducting, then there will be a complicated distribution of charge on the surface, leading to a different force law. That is where JD Jackson starts the treatment of conductors in chapter 3 $\endgroup$ Commented May 8, 2023 at 2:30
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$\begingroup$ @naturallyInconsistent I edited my answer $\endgroup$ Commented May 8, 2023 at 19:06