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### Topic: Parallel-Plate Capacitor Question  (Read 27484 times)

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#### HighTek ##### Parallel-Plate Capacitor Question
« on: March 09, 2009, 12:47:20 PM »
Howdy, I have a problem that I'm a bit confused about. Can someone clarify it for me? Here is the problem:

A parallel-plate capacitor has a plate area of 0.2 m^2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 10^6 V/m between the plates, the magnitude of the charge on each plate should be:

A) 8.9 x 10^-7 C
B) 1.8 x 10^-6 C
C) 3.5 x 10^-6 C
D) 7.1 x 10^-6 C
E) 1.4 x 10^-5 C

Since the problem is asking for charge, I used the equation E=q/Eo*A. Rearranging for q gives me q = Eo*A*E where,

E = Electric field = 2.0 x 10^6 V/m
q = charge
Eo = 8.845 x 10^-12 C^2/Nm^2
A = area = 0.2 m^2

So, q = (8.45x10^-12)*(0.2)*(2x10^6) = 3.5 x 10^-6 C (dimensional analysis works).

So, the answer appears to be C.

However, the question asks for THE MAGNITUDE OF THE CHARGE ON EACH PLATE. Does that mean I now divide 3.5 x 10^-6 C by 2 to get the answer of:

B) 1.8 x 10^-6 C (its actually 1.769 x 10^-6 C)??
I'm experiencing both Alzheimer's and Deja Vu right now, I think I forgot this before.

#### ARGOS++

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« Reply #1 on: March 09, 2009, 01:11:05 PM »

Dear HighTek;

IMHO: If it really asks for Q on each plate, then division by 2.0 is ok.

Think about to create with both plates a new capacitor by combining it with a new neutral plate of same size and identical distance. If you not divide it by 2.0 you would be able to generate this way additional charge/"capacity"/energy.

I hope to have been of help to you with this picure.
Good Luck!
ARGOS++

#### HighTek ##### Re: Parallel-Plate Capacitor Question
« Reply #2 on: March 09, 2009, 02:27:31 PM »
I'm not sure what you mean. I'm am completely lost. I took another approach:

Electric field = Voltage/distance, using the same values given in the original post:

E = V/d => 2.0x10^6 = V/.0001. Solving for V gives me V = 200.

In a second equation, Capacitance = Permitivity constant x (area/distance) or
C = Eo*(A/d).

C = 8.85x10^-12 * (0.2/0.0001).  So..

C = 1.77x10^-8.

Now C also equals charge / voltage or C = q/V. Solving for q gives me q = CV.

q = 1.77x10-8 * 200
q = 3.5x10^-6

Is THAT the magnitude of charge FOR EACH PLATE?
I'm experiencing both Alzheimer's and Deja Vu right now, I think I forgot this before.

#### ARGOS++

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« Reply #3 on: March 09, 2009, 03:43:04 PM »
Dear HighTek;

I used the rearrangement of the following formula:
http://en.wikipedia.org/wiki/Capacitor#Parallel_plate_model
It’s qualitative the identical formula of your second calculation and I got also Q = 3.5416 10-6 C.
But that’s still the charge for the whole capacitor as before, and if you divide the capacitor in two new, as I drew the picture, then you cannot have two with still the whole charge (3.54 10-6 C)  for each.
Think about the law of conservation of energy.

Maybe that other can draw an even better picture for you than I.
Good Luck!
ARGOS++

#### HighTek ##### Re: Parallel-Plate Capacitor Question
« Reply #4 on: March 10, 2009, 04:54:12 PM »
Some have found the answer to be twice the amount of q.

Oh well...
I'm experiencing both Alzheimer's and Deja Vu right now, I think I forgot this before.

#### HighTek ##### Re: Parallel-Plate Capacitor Question
« Reply #5 on: March 13, 2009, 12:32:26 PM »
Well, we got our worksheet back today.  Everyone who answered D) 7.1 x 10^-6 C got it right. Everyone who answered C) 3.5 x 10^-6 C got it wrong.

Naturally, we asked.  He said that in dealing with plate capacitors as a whole, the total charge would have to be zero.

However, when dealing in area, you would need to multiply it by 2 because there are 2 plates.

Then, he said to find the answer to the original post, the equation is q = Eo*A*E, which would be C) 3.5 x 10^-6 C.

Everyone who picked D is now wrong and everyone who picked C is getting extra points.

Its B.S.....I is confused.

So what is the final answer?

I'm experiencing both Alzheimer's and Deja Vu right now, I think I forgot this before.

#### ARGOS++

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« Reply #6 on: March 13, 2009, 01:36:26 PM »

Dear HighTeck;

I’m so sorry that you didn’t got the points!

But I can't believe who somebody is able to tell:
He said that in dealing with ?plate capacitors as a whole?, the total charge would have to be zero.
- if a capacitor "keeps" an electrical field?   That's for me simply impossible!
What would be the sense of a capacitor in this case?  Where is its name coming from?

So my conclusion is only that he asked in real something different, but I have no idea what.
Also about doubling the charge is beyond my grasp.

Its B.S.....I is confused.      So what is the final answer?
So I'm not able to tell you his final answer!