# Electric field of an infinite charged plate

## Homework Statement:

There is an infinite charged plate in yz plane with surface charge density ##\sigma = 8*10^8 C/m^2## and negatively charged particle at coordinate (4,0,0) Find magnitude of efield at coordinate (4,4,0)

## Relevant Equations:

E= E1+E2 So I figured to get e-field at point (4,4,0), I need to find the resultant e-field from the negatively charged particle and the plate
##E_{resultant}=E_{particle}+E_{plate}##
##E_{particle}=\frac{kq}{d^2}=\frac{(9*10^9)(-2*10^-6)}{4^2}=-1125N/C##
Now for the plate is where I'm confused.
If this was a wire, it would have been okay for me since I only need to deal with one dimension.
Since what they requested was a plate in yz plane, does this means that my:
##\sigma=dy*dy*x?## where ##dy## is the 'slice' I take and x is the width of the plate? Is that accurate?
If it is true, then to find the e-field created by that slice at the point, ##dE=\frac{kdq}{R^2}##
##dE=\frac{k\sigma *x*dy}{a^2+y^2}##
I know that the vertical components of the resultant e-field will cancel out because there are same amount of segments on top and below the point.
So need to find ##dE_{x}##, which = ##dEcos\theta##, where ##\theta## is shown: So ##dE_{x} = dEcos\theta = (\frac{k\sigma *x*dy}{a^2+y^2}) (\frac{a}{\sqrt{y^2+a^2}})##,
Now the problem is I can't integrate this to find my resultant e-field because I do not know what the value of x is. If this was a wire in a plane it will have been solvable for me, but now I'm kind of stuck.
Any clues/help? Thanks :)

Related Introductory Physics Homework Help News on Phys.org
TSny
Homework Helper
Gold Member
I'm not sure you are picturing the charged plane correctly. The surface charge density σ is spread on the y-z plane. You do not need to worry about any thickness of the plane. The electric field of a uniformly charged infinite plane is a standard problem. Are you sure you haven't already covered the formula for calculating E due to the plane?

• PeroK
I'm not sure you are picturing the charged plane correctly. The surface charge density σ is spread on the y-z plane. You do not need to worry about any thickness of the plane.
View attachment 249243

The electric field of a uniformly charged infinite plane is a standard problem. Are you sure you haven't already covered the formula for calculating E due to the plane?
Welp didn't noticed this.
Got the formula ##E=\frac{\sigma}{2\epsilon_{0}}=\frac{8*10^{-8}}{2*(8.85*10^{-12})}=4519.77N/C##
Even after adding this with ##1125N/C##, I still can't seem to get my answer right. Any clues?

TSny
Homework Helper
Gold Member
Welp didn't noticed this.
Got the formula ##E=\frac{\sigma}{2\epsilon_{0}}=\frac{8*10^{-8}}{2*(8.85*10^{-12})}=4519.77N/C##
Even after adding this with ##1125N/C##, I still can't seem to get my answer right. Any clues?
Are you using vector addition to add the two electric field vectors?

• PeroK
Are you using vector addition to add the two electric field vectors?
The particle affects the y plane and the plate affects the yz plane?

PeroK
Homework Helper
Gold Member
The particle affects the y plane and the plate affects the yz plane?
You are given the charge distribution in this problem. That is fixed.

kuruman
Homework Helper
Gold Member
The particle affects the y plane and the plate affects the yz plane?
The electric field at point {4,4,0} is the vector sum of two fields: (a) that from the infinite plate and (b) that from the point charge at {4,0,0}. Note that you cannot get a numerical answer unless you have a numerical value for the point charge on the ##x##-axis.

haruspex
Homework Helper
Gold Member
The particle affects the y plane and the plate affects the yz plane?
I don't know what you mean by affecting a plane here. (And there is no "y" plane.)
At the given point, the field from the particle is in the Y direction and that from the plane is in the X direction.

I don't know what you mean by affecting a plane here. (And there is no "y" plane.)
At the given point, the field from the particle is in the Y direction and that from the plane is in the X direction.
The electric field at point {4,4,0} is the vector sum of two fields: (a) that from the infinite plate and (b) that from the point charge at {4,0,0}. Note that you cannot get a numerical answer unless you have a numerical value for the point charge on the ##x##-axis.
The electric field at point {4,4,0} is the vector sum of two fields: (a) that from the infinite plate and (b) that from the point charge at {4,0,0}. Note that you cannot get a numerical answer unless you have a numerical value for the point charge on the ##x##-axis.
Correct me if I'm wrong, so here's how I look at this:
The e-field is at (4,4,0),
the particle can only affect it's y axis (since it has the same coordinates for x and z axis, and the plate can affect its x axis?

TSny
Homework Helper
Gold Member
Correct me if I'm wrong, so here's how I look at this:
The e-field is at (4,4,0),
the particle can only affect it's y axis (since it has the same coordinates for x and z axis, and the plate can affect its x axis?
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
Homework Helper
Gold Member
the particle can only affect it's y axis (since it has the same coordinates for x and z axis, and the plate can affect its x axis?
Is that different from what I wrote in post #8?

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?
Is that different from what I wrote in post #8?
So ##-1125N/C## is in y axis , while ##4519.77N/C## is in x and z axis? So if I want to find the magnitude will it be summing the square of 1125, 4519, 2519 and rooting them?

PeroK
Homework Helper
Gold Member
So ##-1125N/C## is in y axis , while ##4519.77N/C## is in x and z axis? So if I want to find the magnitude will it be summing the square of 1125, 4519, 2519 and rooting them?
The electric field at a point is a vector quantity. To add electric fields you use vector addition. The magnitude is the magnitude of a vector, which you can calculate from its components in the normal way.

You need to be careful that there is not now a gulf between the physics you are studying and your grasp of mathematics.

You should at least try to use proper mathematical terminology.

The electric field at a point is a vector quantity. To add electric fields you use vector addition. The magnitude is the magnitude of a vector, which you can calculate from its components in the normal way.

You need to be careful that there is not now a gulf between the physics you are studying and your grasp of mathematics.

You should at least try to use proper mathematical terminology.
Ok, so to confirm:
##E_{(4,4,0)}=E_{plate} + E_{particle} = (4519.77\widetilde{x}+4519.77\widetilde{z})N/C +(-1125\widetilde{y})N/C##?

haruspex
Homework Helper
Gold Member
Ok, so to confirm:
##E_{(4,4,0)}=E_{plate} + E_{particle} = (4519.77\widetilde{x}+4519.77\widetilde{z})N/C +(-1125\widetilde{y})N/C##?
Why a field in the z axis?

Why a field in the z axis?
Oh.. I thought the plate will also cause efield in the z axis. I'm not picturing the plate right :(

haruspex
Homework Helper
Gold Member
Oh.. I thought the plate will also cause efield in the z axis. I'm not picturing the plate right :(
What is the direction, relative to the plate, of the field from a uniformly charged infinite plate?

What is the direction, relative to the plate, of the field from a uniformly charged infinite plate?
Y and x axis? Towards the point

PeroK
Homework Helper
Gold Member
Y and x axis? Towards the point
No. See the diagram in post #2. Or, look it up. It's a standard configuration.

haruspex
Homework Helper
Gold Member
Y and x axis? Towards the point
Forget the point and any axes for the moment. What direction relative to the plate?

• PeroK
Forget the point and any axes for the moment. What direction relative to the plate?

haruspex
Homework Helper
Gold Member
It's infinite. There is no around.
Which way would a point charge with the same sign as the plate's charge move?

It's infinite. There is no around.
Which way would a point charge with the same sign as the plate's charge move?
Repel, since they have the same charge

haruspex
Homework Helper
Gold Member
Repel, since they have the same charge
Yes, but in exactly what direction relative to the plate?

Yes, but in exactly what direction relative to the plate?
Not sure what you meant, but won't it repel away from its point? for eg: 