# Four metal plates close to each other

1. Jun 28, 2012

### issacnewton

1. The problem statement, all variables and given/known data

Hi
This is a problem from problem book by russian author, Igor Irodov.
Four large metal plates are located at a small distance d from one another as shown in the figure. The extreme plates are connected by a metal wire, while the potential difference V is applied to internal plates. Find

a)the values of electric field strength between neighboring plates

b) the surface charge density on each plate.

2. Relevant equations

$$V_a - V_b= \int_a^b \overrightarrow{E}\circ \overrightarrow{dl}$$

3. The attempt at a solution

I think I got the solution since my answer matched with the answer given at the back of the book. But I have some questions.

First I imagined just plates 2 and 3 connected to a battery. Now battery makes sure that the
two plates will remain at a given potential difference. So now if we bring plates 1 and 4
close to 2 and 3 (which is basically a capacitor), the potential difference between 2 and 3 will not change. And since the problem mentions that 2 and 3 have potential difference of V in the new configuration, the potential difference between 2 and 3 was V in the beginning , before we decided to bring the plates 1 and 4 close to them. Now here I am assuming that as we bring 1 and 4 closer to 2 and 3, the capacitance of 2 and 3 also does not change. I don't know if this assumption is correct though. So with this assumption, since $C=q/V$, and since C and V are constants, charge q will also not change as bring plates 1 and 4 closer.

Now by the symmetry of the problem, E between 1 and 2 will be same as E between 3 and 4. Lets call this E1. Let E between 2 and 3 be E2. Now since 1 and 4 are at the same potential, if point a is on 1 and point b is on 4, then $V_a - V_b =0$. But

$$V_a - V_b= \int_a^b \overrightarrow{E}\circ \overrightarrow{dl}$$

where path goes through 2 and 3 and its straight. So we get

$$0=-E_1 d+V -E_1 d$$

which means $E_1=\frac{V}{2d}$. Also $V= E_2 d$, so we get

$$E_1=\frac{V}{2d},\;\;E_2= \frac{V}{d}$$

Now let $\sigma_1 , \sigma_2$ be the magnitudes of the surface charge densities
on plate 1 and plate 2. Using the fact that electric field due to surface charge density $\sigma$ on a metal plate is given by $E=\frac{\sigma}{2\epsilon_o}$
on both the sides of the plane, I derived the surface charge densities on plates. I got

$$\sigma_1=\frac{\epsilon_o V}{2d},\;\;\sigma_2=\frac{3 \epsilon_o V}{2d}$$

Ok, my answers are correct. But I have question. I made the assumption that as we bring plates 1 and 4 closer to 2 and 3, the capacitance of 2 and 3 remains constant. Is that valid ? and why ? Wont bringing some conductors closer to some capacitor change its capacitance ? Because if we insert a dielectric between the capacitor plates, its capacitor changes.
So it may happen in the situation of this problem.

thanks

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2. Jun 28, 2012

### TSny

Your derivation of E1 and E2 looks good to me. Note that your derivation for the fields does not require you to assume that the capacitance of the system remains constant as you bring in the outer plates.

Once you have E1 and E2 you can determine the charge densities on each face of each plate using E=σ/εo at the surface of a plate.

For example, you should find different charge densities on the top and bottom surfaces of plate 2.

You will see that the total charge on plate 2 is greater with the outer plates in place than if they were not there. So, the capacitance of the system increases when you bring in the outer plates.

3. Jun 28, 2012

### issacnewton

tsny, does the total charge on plate 2 increase because of induction ?. I would think so.
The charge density on plate 1 does not equal the charge density on plate 2 in magnitude because of repulsion from the charges on plate 3, right ?

4. Jun 28, 2012

### TSny

I think that's one way to look at it. My intuition is really not very good. Maybe someone else can provide insight.

Well, the magnitude of the charge density on the lower surface of plate 1 equals the magnitude of the charge density on the upper surface of plate 2 because the magnitude of E is the same at these two surfaces.

But the total charge on plate 1 is less than the total charge on plate 2. I'm not sure I would say that the reason is because of repulsion from plate 3, but, again, my intuition is not good here. Maybe it's ok to think of it that way.

What I find interesting is the following. Suppose you start with just plates 2 and 3 connected to the battery. Then you would have an electric field only between these two plates and essentially no electric field above plate 2 or below plate 3 (assuming very large plates and small separation between the plates). If you now bring in plate 1 from above and plate 4 from below but do not yet connect them together with a wire, then I don't think there would be any induced charge on plates 1 and 4. That's because there is no electric field at their location to induce any charge. But plate 1 would be at a higher potential than plate 4 because if you choose a vertical path from 1 to 4 you will pass through the electric field between plates 2 and 3. Moreover, the charge on plates 2 and 3 would not yet have changed.

When you then connect plates 1 and 4 with a wire they will come to the same potential by having electrons move from plate 4 to plate 1. The resulting negative charge on plate 1 then induces an equal positive charge on the upper surface of plate 2 while the charge on the lower surface of plate 2 remains the same. This extra charge added to plate 2 can be thought of as coming from the battery. Likewise, negative charge is induced on the lower surface of plate 3 from the presence of the positive charge on plate 4.

5. Jun 28, 2012

### Yukoel

Hello isaacnewton,
You can revisualise your arrangement as capacitors connected in parallel.Let me explain. How many ways are there to reach the potential difference of V here ? Plate 2 to 3 (because they have been connected to the battery as so ) /*termed as upper */and plate 2 to 1 to 4 to 3 (V(1)=V(4))/* termed as lower */; so in your case you have a capacitor (2-3 ) parallel to the system of capacitors (2-1 and 4-3 connected in series)This deals away with your problems if you notice that the upper caps (2-3) yield the charge densities on the inner parts and those on the lower branch yield that on the outer surface (of the common plates i.e. 2 and 3).
Now you know that capacitance of plates varies with distance .So isn't bringing 2 and 3 closer gonna change the capacitance of both the upper and lower branches?Upper because the distance is decreasing ,lower because the same is increasing) .Capacitance changes and so does the charge densities .Does this help?
regards
Yukoel

6. Jun 29, 2012

### issacnewton

Yukoel, that makes sense, but think about the original configuration. Initially we just have the capacitor 2-3, there is positive charge on 2 and negative on 3... As we know, there is no electric field outside the capacitor plates,as it cancels out. So if we bring plates 1 and 4 closer from two sides, why would there be any charge accumulation on plates 1 and 4 since there is no electric field where they are located......I have managed to confuse myself....lol....

7. Jun 29, 2012

### Yukoel

Hello isaacnewton,
Just from your side of argument I make a side assumption that the distance between 1-2 and 4-3 is reduced equally(just for avoiding mathematics however little) .The capacitance of the lower branch (as I mention in my last post) still consists of two equal capacitances ,so that the potential difference across each of them remains the same .Now inside a capacitor (In case of large plates) Electric field is supposed to be uniform.Given the new distance ,would the electric field value required for the required potential difference remain the same?Would a new electric field value not require a new charge density?
The result is trivial if they are moved unequally.
Maybe you used the same understanding in this equation

Here different values of E and d may arise for unequal movements and different value of the variable d in case of new configuration.(equal displacement of plates)
Does this help?

regards
Yukoel