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The electric field of two plates

  • Thread starter mborn
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Another question for Gauss' Law;

A very large, thin, flat plate of Aluminium of area A has a total charge Q uniformly distributed over its surfaces. If the same charge is spread uniformly over the upper surface of an otherwise identical glass plate, compare the lectric fields just above the centers of the upper surface of each plate.

M B
 
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Hi,

my answers are;
For Al [tex]E= \frac{\sigma}{\epsilon_o}[/tex]
for Glass [tex]E= \frac{\sigma}{2\epsilon_o}[/tex]

Someone tell me if I am wrong, please
 

Andrew Mason

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hhegab said:
Hi,

my answers are;
For Al [tex]E= \frac{\sigma}{\epsilon_o}[/tex]
for Glass [tex]E= \frac{\sigma}{2\epsilon_o}[/tex]

Someone tell me if I am wrong, please
What about Q and A?

The charge Q spreads out over both surfaces of the aluminum and there is 0 charge inside the metal. So [itex]\sigma = Q/2A[/itex]

The field of the Al sheet is [tex](E_{top} + E_{bot}) = \frac{Q/2A}{\epsilon_0}[/tex]

So: [tex]E_{top} = \frac{Q/2A}{2\epsilon_0}[/itex]

For the glass plate, Q is distributed over area A (not 2A), so [itex]\sigma = Q/A[/itex]

[tex]E_{top} = \frac{Q/A}{2\epsilon_0}[/itex]

AM
 

Gokul43201

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hhegab's right...though I wish he hadn't spilled the solution out.

AM, I believe you have it backwards. In both cases, the surface charge density is usually defined as [itex]\sigma = Q/A [/itex]
For Al :
[tex]\vec{E} = \vec{E1} + \vec{E2} = \sigma /2 \epsilon _0 + \sigma /2 \epsilon _0 = \sigma /\epsilon _0 [/tex]
For glass, there's only the one charged surface.
 

Andrew Mason

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Gokul43201 said:
hhegab's right...though I wish he hadn't spilled the solution out.

AM, I believe you have it backwards. In both cases, the surface charge density is usually defined as [itex]\sigma = Q/A [/itex]
For Al :
[tex]\vec{E} = \vec{E1} + \vec{E2} = \sigma /2 \epsilon _0 + \sigma /2 \epsilon _0 = \sigma /\epsilon _0 [/tex]
For glass, there's only the one charged surface.
I am not sure what you are referring to as being backwards.

[itex]\sigma = Q/2A[/itex] for Aluminum since the charge is distributed over both surfaces of the sheet. For glass [itex]\sigma = Q/A[/itex] since, as you point out, there is only one charged surface of area A.

Just substitute these values for [itex]\sigma[/itex] into your equation to get my answers.

AM
 
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Peace,

In Serway's Physics for S&E ,6th edition, chapter 24 section 4 (conducing surfaces), when using Gauss' Law
[tex] \oint{\vec{E} \cdot d\vec{A}} = \frac{Q_{enc}}{\epsilon_0} [/tex]
is used, we have A as the TOTAL surface area not a surface of a particular region on the material under study. This is how I reached my answer, what do you think?

hhegab
 

Andrew Mason

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hhegab said:
Peace,

In Serway's Physics for S&E ,6th edition, chapter 24 section 4 (conducing surfaces), when using Gauss' Law
[tex] \oint{\vec{E} \cdot d\vec{A}} = \frac{Q_{enc}}{\epsilon_0} [/tex]
is used, we have A as the TOTAL surface area not a surface of a particular region on the material under study. This is how I reached my answer, what do you think?
So what is your answer in terms of Q and A?

AM
 
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[tex] \sigma = \frac{Q}{A} [/tex] in both cases;

hhegab
 

Andrew Mason

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hhegab said:
[tex] \sigma = \frac{Q}{A} [/tex] in both cases;

hhegab
So how does that occur if the surface charge density of the top surface of the alluminum conductor is the same as the bottom surface and there is 0 field in between the two surfaces?

AM
 
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Peace!

The field lines are perpendicular to the surface, and directed either inward or outward according to the sign of sigma, when they added (vectirially) they will give a zero field.
Also, even in the presence of a field inside the conductor, when we put it an an external elecric field, the free charges will arrange themselves on its surface in such a configuration that after about 10^-16 sec. they will cancell the originl field.

hhegab
 

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