# 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?
jisbon
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 :)

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
TSny said:
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?

jisbon said:
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
TSny said:
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?

jisbon said:
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.

jisbon said:
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.

jisbon said:
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.

haruspex said:
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.
kuruman said:
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.
kuruman said:
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?

jisbon said:
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?

jisbon said:
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?

TSny said:
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:
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?

jisbon said:
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.

PeroK said:
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##?

jisbon said:
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?

haruspex said:
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 :(

jisbon said:
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?

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

jisbon said:
Y and x axis? Towards the point
No. See the diagram in post #2. Or, look it up. It's a standard configuration.

jisbon said:
Y and x axis? Towards the point
Forget the point and any axes for the moment. What direction relative to the plate?

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

jisbon said:
It's infinite. There is no around.
Which way would a point charge with the same sign as the plate's charge move?

haruspex said:
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

jisbon said:
Repel, since they have the same charge
Yes, but in exactly what direction relative to the plate?

haruspex said:
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:

jisbon said:
Not sure what you meant, but won't it repel away from its point? for eg:

View attachment 249633
It is an infinite plate. Why would the repulsion be down (in terms of your diagram) rather than up?

jisbon said:
Not sure what you meant, but won't it repel away from its point? for eg:

View attachment 249633

Pardon for messy drawing, but if I'm correct, the charge particle simply repels to the right right? Since the vertical components are pretty much canceled out

Delta2 and PeroK
Seems like my idea was correct. However, in my initial attempt, I have already understood that vertical forces will be canceled. So hence my resultant force:
force from plate (only affects x axis) + force from particle (only affects y axis)?

jisbon said:
Seems like my idea was correct. However, in my initial attempt, I have already understood that vertical forces will be canceled. So hence my resultant force:
force from plate (only affects x axis) + force from particle (only affects y axis)?
Yes, but in post #14 you had the plate generating a field in the z axis also.

haruspex said:
Yes, but in post #14 you had the plate generating a field in the z axis also.

##E_{(4,4,0)}=E_{plate} + E_{particle} = (4519.77\widetilde{x})N/C +(-1125\widetilde{y})N/C##?

jisbon said:

##E_{(4,4,0)}=E_{plate} + E_{particle} = (4519.77\widetilde{x})N/C +(-1125\widetilde{y})N/C##?
Yes.

haruspex said:
Yes.
I've calculated the magnitude and it appears to be 4657, which is different from the given answer, which is 9109 :/

I think your answer is correct if the infinite plate has only one face with charge density ##\sigma##. However, if they meant the plate has two faces, each face having charge density 8 x 10-8 C/m2, then you would get their answer. If this latter interpretation is what they meant, then I misled you with the diagram in post #2. Sorry about that.

Last edited:
TSny said:
I think your answer is correct if the infinite plate has only one face with charge density ##\sigma##. However, if they meant the plate has two faces, each face having charge density 8 x 10-8 C/m2, then you would get their answer. If this latter interpretation is what they meant, then I misled you with the diagram in post #2. Sorry about that.
Hmm okay. So if it has two faces, is there now supposed to be a negative x direction now? Not sure how to proceed

jisbon said:
Hmm okay. So if it has two faces, is there now supposed to be a negative x direction now? Not sure how to proceed
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.

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