Electric field of an infinite charged plate

[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 :)

 

[TSny]

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?

 

[jisbon]

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]

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?

 

[jisbon]

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]

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]

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]

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.

 

[jisbon]

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]

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]

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?

 

[jisbon]

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]

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.

 

[jisbon]

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]

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?

 

[jisbon]

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]

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?

 

[jisbon]

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]

Y and x axis? Towards the point

No. See the diagram in post #2. Or, look it up. It's a standard configuration.

 

[haruspex]

Y and x axis? Towards the point

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

 

[jisbon]

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

Spreading around the plate?

 

[haruspex]

Spreading around the plate?

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

 

[jisbon]

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]

Repel, since they have the same charge

Yes, but in exactly what direction relative to the plate?

 

[jisbon]

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:

 

[haruspex]

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]

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

 

[jisbon]

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)?

 

[haruspex]

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.

 

[jisbon]

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

Oh.. my bad. So its:

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