Charge Distribution in a Parallel Plate Capacitor with Connected Plates

[carlyn medona]

Homework Statement

Each plate in the figure has area a and separation d , end of battery is e find charge supplied by battery

Homework Equations

The Attempt at a Solution

. each facing plate will have half the charge supplied by battery, but I am confused how the potential is applied[/B]

 

[haruspex]

each facing plate

Do you mean the two connected to the battery? If so...

will have half the charge supplied by battery

... but they are connected to opposite poles.

For the two plates not connected to the battery, what is the relationship between their charges, and what is the relationship between their potentials?

 

[carlyn medona]

I mean magnitude of charge on each facing plate, won't that be q÷2 and, the plates connected together are at same poyential

 

[haruspex]

each facing plate

I still do not know which plates you mean. There are four plates, and they face each other.

won't that be q÷2

Where q is what?

the plates connected together are at same poyential

Yes.

 

[carlyn medona]

Q is charge supplied by battery

 

[haruspex]

Q is charge supplied by battery

Does the battery gain a net charge?

 

[rude man]

You have 3 capacitors in series. Draw a schematic diagram, simplify if possible, then solve as a circuit problem.

 

[carlyn medona]

Is potential difference across the series connection equal to the emf of the battery?

 

[rude man]

Is potential difference across the series connection equal to the emf of the battery?

No. Did you draw the diagram?

 

[haruspex]

@carlyn medona , you did not answer my question. You say q is "the charge supplied by the battery", which is fine, but seem to think that means it supplies q/2 to a plate on each terminal, which is definitely not fine. That would mean the battery itself acquires a net charge of -q, which is nonsense.
The terminals are oppositely in sign, so what does a battery "supplying a charge of q" mean?

 

[carlyn medona]

Okay, I thought this is how charge gets distributed, please correct me if I am wrong

 

[rude man]

Okay, I thought this is how charge gets distributed, please correct me if I am wrong

Your first tipoff that something is wrong is as follows: put a test charge inside the leftmost plate (4). The E field in that plate (E4) is of course zero. Now look at the contributions to E4 from all 8 sides of charge.
From side 8 (left side of leftmost plate) you get some E field, if non-zero, pointing, say, to the left (- charges on side 8). Now, sides 7 thru 2 cancel out any contribution to E4 since those are pairs of equal and opposite charge. That leaves side 1 to contribute to E4. But surface charge density on side 1 must be equal to that of side 8 in order to get a zero E4 field, yet you have those sides as being of opposite polarity, so something is wrong.

You'll have to show your work for further help. You do have equal and opposite charges on all inside adjoining surfaces, and you have Q2+Q4=0, and Q1+Q2+Q3+Q4=0, so that 's good. But somewhere you made a slipup (assuming I'm right to begin with ).

 

[haruspex]

Okay, I thought this is how charge gets distributed, please correct me if I am wrong

At last I know what you mean by "q/2 to each facing plate". You meant q/2 to each face of one plate (and -q/2 to each face of the other plate). A pity you did not try to explain that before.

I suggest you start by ignoring the other two plates and consider how charge would be distributed across the faces of the two connected to the battery. Would you still put magnitude |q/2| on each face? Then insert the other two plates and consider how charge would rearrange itself on those. Of course, that might lead to rearrangement of charge on the first two, but it should give you a better understanding of how things should look.

 

[carlyn medona]

I got electric field inside each plate to be zero, but not sure about charge distribution

 

[rude man]

I got electric field inside each plate to be zero, but not sure about charge distribution

This is incomplete and hard to read.
Show your work.

BTW if you haven't covered capacitance yet you can derive all 8 charge densities by the two equations in Q1-Q4 I posted in #12, plus σ = D plus V = - d⋅E.

(σ = Q/area, D = εE).

 

[haruspex]

I got electric field inside each plate to be zero, but not sure about charge distribution

The other thing to check is whether the two plates connected to each other are at the same potential. If so,that must be the solution.

 

[rude man]

The other thing to check is whether the two plates connected to each other are at the same potential. If so,that must be the solution.

That's true, but consider: you have 6 unknowns (the 5 non-redundant side charge densities plus the potential of the connected two plates). So you need 6 independent equations to solve for everything. Try to use the hints I gave you last time to come up with these equations.