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Old May 22nd 2009, 07:47 PM   #1
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Potential drop

The potential displacement graph is as shown in the attachment. The red line shows the potential when the metal block is not inserted. The black line shows potential when it is inserted. My teacher tells me that the drop in voltage when displacement is 0 m is due to the induced negative charges. But when we consider the other side, there is positive induced charges, why aren't there a drop in potential?
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Potential drop-potential-difference.gif  
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Old May 22nd 2009, 09:00 PM   #2
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The negative induced charge is closer to the positive plate of the capacitor, hence it decreases the potential of the positive plate of the capacitor. The positive induced charge force increases the potential of the positive plate but being farther its effect is less than that due to negative induced charge.
Like wise positive induced charge being closer to negative plate of the capacitor, increases the potential of the negative plate.
Now potential of positive plate of the capacitor decreases and that of the negative plate increases and hence the potential difference between the two plates decreases.
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Old May 23rd 2009, 09:59 PM   #3
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Why aren't the shape like this?

Even if the metal conductor, the induced charge would affect the potential across metal plates?

Thanks
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Potential drop-pd.gif  
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Old May 24th 2009, 09:12 AM   #4
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Originally Posted by werehk View Post
The red line shows the potential when the metal block is not inserted.
I am probably not understating the question well. When there is no metal plate inserted then in the potential - displacement graph (red line ) what is displaced ?
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Old May 25th 2009, 12:01 AM   #5
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Sorry that I draw the wrong picture last time. Should be this one.Potential drop-pd2.gif

Metal plates are present with a difference in voltage originally. And the potential difference across the plates are shown by the red line. Now, insert a metal block and there shows a sudden drop in voltage on one side of the plate as there is induced charges inside the metal block.(shown by the black line of my first picture)

But both sides of the metal block has got induced charges, why aren't there drop in voltage on both sides of the plates?(shown by the black line of picture just uploaded)

Last edited by werehk; May 25th 2009 at 12:11 AM.
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Old May 25th 2009, 10:16 AM   #6
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(once again ) at +10 V and negative one at -10 V we say that positive plate is at at a potential of 20 V wrt negative plate( means considering negative plate at zero potential). If now a metal block is inserted, the
I think I got it now.
The potential of the positive plate is measured wrt negative plate. Potential V is actually the potential difference between negative and positive plate. Absolute potential has no meaning. For example if negative plate is at -10 V and the positive at +10 V OR the negative plate is at 0 V and positive one at 20 V, the two statements are same. If the negative plate is not earthed, inserting metal block decreases the potential of positive plate and increases the potential of negative plate thus overall decreasing the potential difference between the two plates. For example before inserting the metal block the positive plate is potential of positive plate decreases to, say +8 V and that of negative plate increases to, say -8 V and hence the potential difference is now 16 V. We say that positive plate has a potential 16 V wrt negative plate ( again considering negative plate at zero )

Now coming to your question.

In absence of the metal block:

As you move from positive to negative plate, the potential decreases uniformly (V = Ed) from V at positive plate to zero at negative plate and hence the red line of the graph.

When the metal block is inserted:

As explained earlier, due to induced charge the potential difference between the two plates decrease suddenly. Now if you move from positive to negative plate, the potential up to left side of metal block decreases uniformly and the graph has a negative slope. The metal block is an equipotential volume( field inside the metal block is zero).Hence the potential remains constant in side the metal block and hence the flat part of the graph. Now when you further move from right side of the metal block to the negative plate of the capacitor, potential again decreases uniformly.
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