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
Net electric field on negatively charged particle in front of infinite charged sheet
electric-fieldselectrostatics
Related Solutions
It sounds like the math is not the problem. It is visualizing the geometry.
A point charge fills all space with an electric field. If you have multiple charges, they all create electric fields. To find the electric field at a point, add them up.
Do this with a sphere. If you are outside as shown above, all the E field contributions are more or less to the right. It works out just the same as a point charge E field. If you are inside, contributions from the left and right cancel perfectly. The total is 0. This property of the inverse square law has always struck me as one of the coolest things in physics.
With two spheres, the inner point has 0 field because contributions from the red sphere cancel and so do contributions from the blue sphere.
For the middle point, the red sphere adds up to the same as a point charge, but the blue sphere contributions add to 0.
For the outer point, each sphere adds up to the equivalent of a point charge. In this case two opposite point charges at the same point would cancel.
For a plane, by symmetry the contributions add up to an E field perpendicular to the plane. The plane is infinite, and so the charge is infinite. But some is infinitely far away. The E field works out to be finite and uniform.
For two oppositely charged planes you get 0 on the outside and reinforcement in between. This is indeed similar to the two spheres, even though there is no charge off to the left.
Imagine the two spheres getting larger and larger, so the geometry looks more and more like two planes. The E field should approach the same solution as the two planes.
The left hand point is always inside two spheres. The right hand point is outside of two large spheres that always cancel. So the field is always 0 for them.
For the middle point, the red sphere matters. For the spheres to become like the planes, we must keep the charge density constant as they grow. This means the total charge grows larger. The total red E field is equivalent to a larger, but more distant point charge.
If the radius is r, the total charge, Q, is proportional to $r^2$, and the field is proportional to $Q/r^2$. So the field doesn't change. The E field becomes more uniform, and more like the plane.
The difference between the finite and the infinite sheet, is that with the infinite sheet, we can assume by symmetry that $\vec{E}$ is everywhere perpendicular to the surface, which makes taking the surface integral for Gauss's law trivial.
When the sheet is finite, we can no longer make that assumption, and would need to consider edge effects. Indeed as you move further away from the finite plane, the smaller and smaller it will look, eventually if you were far enough away, you could imagine it would look almost like a point charge, and you'd expect a $\frac{\vec{r}}{r^3}$ type of field in the far limit. Intuitively, with a finite sheet, you could imagine looking at the size of the sheet to say how far you are away from the sheet. But with the infinite sheet, no matter how far you are away from it, you only ever see an infinite sheet. So the distance away from the sheet has no meaning, hence the field must be constant at all distances.
Best Answer
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.