Charge Distribution on Metal Sheet

[Saitama]

no, i meant taking advantage of one of the basic symmetries of the whole set-up

(and remember, when you have a conductor, part of your gaussian surface is likely to be along the middle of it)

you might get a clue by working backwards from the answer … what do you need the charges to be?

But if I select a gaussian surface in the middle, far from both the ends, it won't include the charge Q/2. I am honestly lost.

It's also possible to "brute force" the solution if you know the equation for the surface charge distribution induced on an infinite flat sheet by a point charge q located at a distance h from the sheet. This would involve doing some integration. This method gives the Q/4 given results you alluded to earlier (PA), but, with the opposite signs.

I am very sorry for the confusion, the given answers are indeed opposite. Sorry again.

But how do you approach it through integration?

 

[Chestermiller]

But how do you approach it through integration?

The equation for the surface charge distribution induced on an infinite flat sheet by a point charge q located at a distance h from the sheet is:
$$σ(x,y)=-\frac{q}{2π}\frac{h}{(x^2+y^2+h^2)^{3/2}}$$
where x in this equation is measured from the point directly underneath the charge. (http://www.science.org.ge/moambe/2007-vol2/bolotov.pdf)

When you have a charge of -Q/2 located at (-h,0,h) and a charge of +Q/2 located at (h,0,h), and the potential at x = 0 is equal to zero (as in our situation), you can treat the two semi-infinite sheets as a single infinite sheet. The charge distribution due to the +Q/2 charge located at (h,0,h) is given by:
$$σ(x,y)=-\frac{Q}{4π}\frac{h}{((x-h)^2+y^2+h^2)^{3/2}}$$
and the charge distribution due to the -Q/2 charge located at (-h,0,h) is given by:
$$σ(x,y)=+\frac{Q}{4π}\frac{h}{((x+h)^2+y^2+h^2)^{3/2}}$$
You just superimpose these two charge distributions and integrate over the area of the right plate (y=-∞ to y=+∞, x = 0 to x = ∞) to get the overall charge on the right plate. It should come out to -Q/4.

Chet

 

[BvU]

Well, you manage to get some hefty assistance for this cute problem. As Chet started up and Tsny and Tim brought to an end, there is a high-brow basic trickery that indeed can be pulled off here. I won't spoil T&T's fun by telling.

In textbook exercises like this, d horizontal conveniently and coincidentally happens to be equal to d vertical; never occurs in practice. I didn't even bother to explore it (meaning: kudos to T&T for making good use of it).

Coming back to my post #2 and your asking how to set up the integral in post #5:
The thing I liked was that if you put the origin on the right plate right underneath Q, there is a left-right symmetry and you realize that the left plane is the same as the right plane from d to ∞.

So with ∑Q_induced= 0, whatever we find for the center strip (-d to +d) plus twice Q_{induced, left} should give 0.

For point P = P(x,y,0) the z-component of the electric field for z > 0 (there is no E field for z in or below the plate) is $$ E_\perp = {1\over 4\pi \epsilon_0} \ {Q \over r^2 } \ \left (\hat {\rm \bf r}_1 \cdot \hat {\rm \bf k} - \hat {\rm \bf r}_2 \cdot \hat {\rm \bf k} \right ) \quad {\rm with }\enspace r^2 = x^2+y^2+d^2, \enspace{\rm and }\enspace \vec r_1 = (x, y, -d),\enspace \vec r_2 = (x, y, d), $$ so that $$ E_\perp = -{2Q\ d\over 4\pi \epsilon_0\ r^3} $$ use Gauβ $E={\sigma \over 2 \epsilon_0}$, first integrating over x from -d to d, then y from -∞ to ∞ should give -Q/2.
Chet did it, and otherwise check out MIT and UIUC Errede, in particular the helpful integrals in MIT, top p. 34.

Book answer must have sign wrong: opposite +Q a negative charge should dominate.

 

[TSny]

But if I select a gaussian surface in the middle, far from both the ends, it won't include the charge Q/2.

Try to construct a Gaussian surface that encloses the point charge and for which you know the value of E at every point on the surface.

 

[Saitama]

Try to construct a Gaussian surface that encloses the point charge and for which you know the value of E at every point on the surface.

I can't think of anything which would be useful. Selecting a sphere centred at charge won't be a good idea because the net electric field at each point on sphere is unknown owing to the charges induced on the plate. Tiny-tim said no to a cylinder. Cube probably won't work too.

I didn't really expect that this problem would get so complicated as this problem is from a high school level problem set. I thought I was missing something basic so decided to post here but looking at the above posts, I guess I was wrong. :(

 

[TSny]

I can't think of anything which would be useful. Selecting a sphere centred at charge won't be a good idea because the net electric field at each point on sphere is unknown owing to the charges induced on the plate. Tiny-tim said no to a cylinder. Cube probably won't work too.

What is the value of E inside the conductor?

Can you argue that outside the conductor and for large distances from the point charge E must fall off with distance from the point charge faster than 1/r²? If so, what does that tell you about the flux through a surface "at infinity"?

Can you construct a closed surface that is partly inside the conductor and partly "at infinity" and encloses the point charge?

I didn't really expect that this problem would get so complicated as this problem is from a high school level problem set. I thought I was missing something basic so decided to post here but looking at the above posts, I guess I was wrong. :(

I'm also surprised that it's a problem in a high school set. Seems to me that it's more at the level of Purcell's Electricity and Magnetism, a college text. But, maybe I'm not seeing a more elementary way of getting the answer!

 

[Saitama]

What is the value of E inside the conductor?

Zero.

Can you argue that outside the conductor and for large distances from the point charge E must fall off with distance from the point charge faster than 1/r²? If so, what does that tell you about the flux through a surface "at infinity"?

I honestly don't see why the field would fall that way in this case. I have seen this in the field of dipole where it falls at 1/r^3.

Can you construct a closed surface that is partly inside the conductor and partly "at infinity" and encloses the point charge?

I can't imagine how that surface could be, I have never selected gaussian surfaces of that kind.

I'm also surprised that it's a problem in a high school set. Seems to me that it's more at the level of Purcell's Electricity and Magnetism, a college text. But, maybe I'm not seeing a more elementary way of getting the answer!

Ah, I do have the book btw (which I bought during my short appearance at college, now I am back to high school stuff ).

Here is the source of problem: http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201401&t=fiz&l=en

 

[Chestermiller]

Can you construct a closed surface that is partly inside the conductor and partly "at infinity" and encloses the point charge?

Hi TSny. The part I'm having trouble with in applying Gauss's law is what to do about the vertical zero potential surface. Even though the component of the electric field tangent to this surface is zero throughout the surface, the electric field is normal to the surface and is a function of position on the surface. Do you have in mind replacing the surface with an equivalent conducive sheet, with positive and negative charges on the opposing surfaces of the sheet? I'm not confident about doing something like that, given my limited experience with electrostatics.

Chet

 

[TSny]

Hi TSny. The part I'm having trouble with in applying Gauss's law is what to do about the vertical zero potential surface. Even though the component of the electric field tangent to this surface is zero throughout the surface, the electric field is normal to the surface and is a function of position on the surface. Do you have in mind replacing the surface with an equivalent conducive sheet, with positive and negative charges on the opposing surfaces of the sheet? I'm not confident about doing something like that, given my limited experience with electrostatics.

Yes, we add a vertically conducting plate as shown in the third figure attached.

In the three figures shown, the electric field in the first quadrant is the same for each figure. In the figure on the left, the charge induced on the portion of the horizontal plate where x > 0 is the same as for the right figure.

Also, due to symmetry in the third figure, you can compare the charge on the vertical plate to the charge on the horizontal plate.

Using Gauss' law you can determine the total charge induced on both the horizontal and vertical plates in the third figure. [EDIT: Note that you can choose parts of the Gaussian surface to lie in the xy horizontal plane and the yz vertical plane where it may be considered to be inside conducting material.]

 

[TSny]

Zero.

Yes.

I honestly don't see why the field would fall that way in this case. I have seen this in the field of dipole where it falls at 1/r^3.

Look at the middle figure that I posted in #44. Far from the charges, would the field be that of a point charge, a dipole, a quadrupole, or something else?

I can't imagine how that surface could be, I have never selected gaussian surfaces of that kind.

A hint is given in the "Edit" of my post #44. Choose part of the surface to be flat and lying in the horizontal xy plane for x>0. Another part will be flat and lying in the vertical yz plane for z>0. You will need to add more parts to make a closed surface and decide how big to make the whole thing.

Ah, I do have the book btw (which I bought during my short appearance at college, now I am back to high school stuff ).

Here is the source of problem: http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201401&t=fiz&l=en

Nice problems. In my opinion, some of these are definitely more advanced than typical high school physics problems in the United States. But then, we all know that Hungarians are from Mars

 

[ehild]

Here is the source of problem: http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201401&t=fiz&l=en

That is Hungarian! I used to solve problems written in that Journal when at high school, more than half century ago... Those problems are very cunning and are for the best students.

I would solve it as follows, see attachment. The charge Q induces -Q surface charge on the upper surface of the plates. The plates are divided into strips, the surface charge densities are q₂ on the left plane, q₁ in the strips symmetric to the charge Q, between x=0 and x=±D, and q₂ on the rightmost strip. q₁+q₂=-Q/2

You get q₁ by integrating ε₀ in the range 0<x<D, -∞<y<∞. Integrate with respect to y first. The following integral is helpful (from Wolframalpha)

$$\int{\frac{1}{(\sqrt{y^2+a^2})^3}dy}=\frac{y}{a^2\sqrt{y^2+a^2}}dy$$

The total charge of the connected plates is zero. Because of the induced charge on the upper surfaces, there is surface charge on the back surfaces, too. The back surface charge is evenly distributed. The plates are of equal size. So the total charge of the left plate is q₂+Q/2 and the charge of the right plate is 2q₁+q₂+Q/2.

ehild

 

[BvU]

@ehild: You agree that the Hungarian "solution" is wrong by a factor of -1 ?

Re your integral: could you show us how to integrate wrt y first ? I find it rather awkward...

 

[Chestermiller]

Yes, we add a vertically conducting plate as shown in the third figure attached.

In the three figures shown, the electric field in the first quadrant is the same for each figure. In the figure on the left, the charge induced on the portion of the horizontal plate where x > 0 is the same as for the right figure.

Also, due to symmetry in the third figure, you can compare the charge on the vertical plate to the charge on the horizontal plate.

Using Gauss' law you can determine the total charge induced on both the horizontal and vertical plates in the third figure. [EDIT: Note that you can choose parts of the Gaussian surface to lie in the xy horizontal plane and the yz vertical plane where it may be considered to be inside conducting material.]

Thanks TSny. I've finally gotten the idea of how to do it. But I would have set it up a little differently than you have done it. With reference to your figures, in the first figure, our starting point is with the charges equal to -Q/2 and + Q/2. I would then have represented this figure as the sum of two other figures: (figure 1). Your middle figure with Q/4 s, and the signs on the charges the same as in your figure, plus (figure 2). Your middle figure with Q/4 s, but with both positive charges on the right and both negative charges on the left.

Because of symmetry, the setup in (figure 1) would not give rise to any net induced charges on the two plates. But, the setup in (figure 2) would give rise to the same total induced charges on the plates as in the original problem. Also, in (figure 2), the charges on the upper and lower surfaces of each plate would be identical, and the electric fields within the plates would be zero. I would then place a vertical plate on the same vertical zero-potential surface that you have used in your third figure. This plate would develop an induced negative charge on its right surface and an induced positive charge on its left surface. The electric field within this plate would be zero, and the charge distribution on the right side of the plate would be identical to the charge distribution on the upper surface of our right plate. This would represent my third figure. I would put one Gaussian planar surface down the middle of the right plate, another Gaussian planar surface up the middle of the vertical plate, and a third Gaussian surface at infinity. So, my third figure would look essentially identical to yours. But the point charge would be +Q/4, only half the charge on the right plate would be included within the Gaussian surface, and only the negative charge on the vertical plate would be included (and this would equal the charge on the upper surface of the right plate). If q was half the charge on the negative plate, then by applying Gauss's law, q = -Q/8. So, the full charge on the right plate would be -Q/4.

I hope this makes some kind of sense. It was very hard to describe in words. If you were able to understand my description, does this work for you?

Chet

 

[TSny]

I hope this makes some kind of sense. It was very hard to describe in words. If you were able to understand my description, does this work for you?

Chet, I'm pretty sure I followed your logic and it looks correct to me. Nice!

 

[ehild]

@ehild: You agree that the Hungarian "solution" is wrong by a factor of -1 ?

Re your integral: could you show us how to integrate wrt y first ? I find it rather awkward...

The solution in Hungarian

A bal oldali lemezen Q/4, a jobb oldalin -Q/4 lesz a töltés.


means

'The charge on the left plate will be Q/4, that on the right plate will be -Q/4.'

(bal =left, jobb = right) The Hungarian solution is correct.

Applying the method of mirror charges, the electric field on the upper surface is normal to the plate, pointing downward, and

$$E_z=-\frac{1}{2\pi\epsilon_0}\frac{Q D }{(x^2 + y^2 + D^2)^{3/2} }$$

The induced charge q1 is the integral of σ=ε₀E_z for the orange strip

$$q_1=-\int_{-\infty} ^{\infty}\int_0^D{\frac{1}{2\pi}\frac{Q D }{(x^2 + y^2 + D^2)^{3/2} }dydx}$$

apart from the constant factors, the integrand is of the form $$\frac{1}{(y^2 + a^2)^{3/2}}$$
with a^2=x^2+D^2.

The integral is shown in my last post.

ehild

 

[BvU]

Excellent, thank you.
Oops, I completely forget Pranav: Do you already see the Gauss volume they are dangling in front of you ? And then a symmetry plane on its 45 degree axis (oops...) ?
That's two oopses in one post. Now it's even three.

 

[Saitama]

Hi ehild!

Great solution but I have a few doubts regarding it.

I would solve it as follows, see attachment. The charge Q induces -Q surface charge on the upper surface of the plates. The plates are divided into strips, the surface charge densities are q₂ on the left plane, q₁ in the strips symmetric to the charge Q, between x=0 and x=±D, and q₂ on the rightmost strip. q₁+q₂=-Q/2

I am sorry if I am missing something but how do you get $q_1+q_2=-Q/2$. You said that the positive charge induced on the surface beneath gets evenly distributed. I think it should be $q_1+q_2=-Q$.

You get q₁ by integrating ε₀ in the range 0<x<D, -∞<y<∞. Integrate with respect to y first. The following integral is helpful (from Wolframalpha)

$$\int{\frac{1}{(\sqrt{y^2+a^2})^3}dy}=\frac{y}{a^2\sqrt{y^2+a^2}}dy$$

Why integrate only for 0 to D only? In your later post, you said that you used the method of images. It is possible to find E at every point on the sheet, right? So why not integrate x from -D to ∞?

One more thing, if this problem is easily solved by method of images, why TSny and Chestermiller are using such complicated methods and I still have zero idea about the Gaussian surface, I feel I am only wasting the precious time of TSny and Chestermiller by asking more and more hints and making zero progress.

 

[ehild]

Hi ehild!

Great solution but I have a few doubts regarding it. I am sorry if I am missing something but how do you get $q_1+q_2=-Q/2$. You said that the positive charge induced on the surface beneath gets evenly distributed. I think it should be $q_1+q_2=-Q$.

see my picture: there are two pairs of strips with induced charge q1 and q2 on the top surface of the two connected plates.The induced charge is -Q. -Q=2q1+2q2.

Why integrate only for 0 to D only? In your later post, you said that you used the method of images. It is possible to find E at every point on the sheet, right? So why not integrate x from -D to ∞?

I felt it easier. I tried to use symmetry. But you can avoid my strips and integrate to the whole plate.

One more thing, if this problem is easily solved by method of images, why TSny and Chestermiller are using such complicated methods and I still have zero idea about the Gaussian surface, I feel I am only wasting the precious time of TSny and Chestermiller by asking more and more hints and making zero progress.

To tell the truth, I do not understand their method, either. They suggest to extend the Gaussian surface to infinity. ehild

 

[Saitama]

I felt it easier. I tried to use symmetry. But you can avoid my strips and integrate to the whole plate.

I think I see why you did that. Simply brilliant.

Thanks a lot ehild!

 

[TSny]

I've made an attempt to summarize Chestermiller's "no integration" method with a couple of pictures. Any errors are my own.

The method is to replace the original problem with a superposition of other problems that are easier to analyze but still produce the same net charge on the right half of the horizontal plate.

The first picture attached shows the original problem as Figure A. This figure is thought of as a superposition of Figure B and Figure C. Since Figure B does not contribute any net charge to the right plate, we can just consider Figure C. Note that the distribution of the net charge on the right plate in Figure C is not the same as the distribution in Figure A. But, that's OK since we are only interested in the net charge on the right plate, and the net charge on the right plate is the same in Figures A and C. (The distribution of charge in Figure A is the superposition of the distributions in Figures B and C.)

In the second picture attached, Figure C is replaced by a superposition of Figure 1 and Figure 2. Figure 1 provides no net charge on the right plate. So, the net charge on the right plate in Figure 2 is the same as the net charge on the right plate in Figure C which is the same as the net charge on the right plate of the original problem (Figure A).

A vertical conducting plate is added to Figure 2 to get Figure 3. This additional plate does not disturb the charge distribution on the horizontal plate. Then, using Gauss' law and symmetry you can deduce the surface charge on the plates in Figure 3 and arrive at the final answer.

 

[Saitama]

Thanks TSny!

In the second picture attached, Figure C is replaced by a superposition of Figure 1 and Figure 2. Figure 1 provides no net charge on the right plate. So, the net charge on the right plate in Figure 2 is the same as the net charge on the right plate in Figure C which is the same as the net charge on the right plate of the original problem (Figure A).

Ok, this is going to be silly, why no charges are induced in the case of figure 1?

A vertical conducting plate is added to Figure 2 to get Figure 3. This additional plate does not disturb the charge distribution on the horizontal plate. Then, using Gauss' law and symmetry you can deduce the surface charge on the plates in Figure 3 and arrive at the final answer.

I still don't get the point of adding the conducting plate and why it doesn't affect the given setup. Is it not possible to find the induced charges directly without the conducting plate? Why does adding the conducting plate allows one to use these symmetrical arguments to reach the answer?

Is this method illustrated in any book? I have both Purcell and Griffiths, is this method present in any of the two books?

 

[ehild]

The method is to replace the original problem with a superposition of other problems that are easier to analyze but still produce the same net charge on the right half of the horizontal plate.

It is ingenious! Thank you TSny.

ehild

 

[TSny]

It is ingenious!

Yes. Hats off to Chestermiller.

 

[ehild]

My only concern is applying Gauss' Law. For that, it had to be known that the electric field tends to zero faster then 1/r² (r is the distance from the charge in the quarter space).
Anyway, the symmetry considerations and superposition principle work well. And that was Chest's idea. ehild

 

[ehild]

Ok, this is going to be silly, why no charges are induced in the case of figure 1?

Charges are induced on the top surfaces of both plates, but they are equal because of symmetry, just like the opposite charge on the back surfaces. So the net charge of both plates is zero.

I still don't get the point of adding the conducting plate and why it doesn't affect the given setup. Is it not possible to find the induced charges directly without the conducting plate? Why does adding the conducting plate allows one to use these symmetrical arguments to reach the answer?

Is this method illustrated in any book? I have both Purcell and Griffiths, is this method present in any of the two books?

An equipotential surface and a metal surface are alike as the potential is the same and the electric field lines are normal to both. You get the same electric field of the four charges in each quadrant without the vertical plate and with it. In the latter case, when determining the field in one quadrant, the mirror charges are the same at the same places as the real ones.

The superposition principle is applied at the very beginning of studies of Electricity. When you determine the field of two charges, you just add the contributions.
And symmetry is very inherent property - if something is symmetric, its properties are also symmetric.

You can not learn every method of solution of problems. You have to practice and then you have the eye to see the solution. And it is no problem if you do not find the nicest method. In this problem, the integral method was also satisfactory.

In real life, you need to solve problems which do not have nice solutions.


ehild

 

[Chestermiller]

Yes. Hats off to Chestermiller.

Thanks TSny. You did a beautiful job of illustrating and articulating what I was trying to say. I might also add that, initially, I was unable to figure out how to do this, and it wasn't until I received the hints from yourself and Tiny Tim that I was able to put this analysis together. So, in reality, this was really a team effort.

Chet

 

[TSny]

My only concern is applying Gauss' Law. For that, it had to be known that the electric field tends to zero faster then 1/r² (r is the distance from the charge in the quarter space).

I think that for the limit of infinite plates, you can argue that the field in the first quadrant for the left figure is the same as in the middle figure where the other three charges have been removed. That's because for infinite plates, the region of the first quadrant alone is a well-defined boundary valued problem. The solution of the field for the first quadrant can then be obtained by the method of images and would be the same field as produced in the first quadrant by the four point charges alone of the figure on the right. This is a quadrupole field that will fall off rapidly with distance.

But, maybe I'm overlooking something.

I've spent about ten times as much time trying to think through this problem without integration as it took to just do the integration. But it's fun.

 

[Saitama]

I am still going through all the replies in this thread, thanks a lot everyone! :)

That was an information overload for me this time.

 

[tiny-tim]

Thanks TSny. You did a beautiful job of illustrating and articulating what I was trying to say. I might also add that, initially, I was unable to figure out how to do this, and it wasn't until I received the hints from yourself and Tiny Tim …

yes, thanks TSny!

for the record, i had no idea until i saw TSny's hint at post #26!

 

[rude man]

Talk about a late entry! But, I'd go with:

1. Consider the two plates as one.
2. Image the charge below the plate at z = -d.
3. Compute the E field all along the plate based on the removal of the plate and its substitution by the (negative) image charge.
4. Use σ = ε₀E to get σ(x,y).
5. Integrate σ to get the charge distribution along the plate, then divide it up per the location of Q.

Bottom line, I would base my approach on the image technique. I don't know to what extent this was done by all the other posters.