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## Main Question or Discussion Point

Hi guys

I have some questions about electrostatics, and I hope you can help me. Here they are:

1) Please take a look at the following example: http://books.google.dk/books?id=YkGQFFhZWicC&pg=PA387&lpg=PA387&dq="electric+field+produced+by+a+uniformly+polarized"&source=web&ots=4pobXLVw66&sig=YiULcbQhTa7htSGbHJ4-D4ZBaJU&hl=da

Here we are calculating the electric field

2) I have two conducting metal plates with a distance

[tex]

V = - \int_C^r {\overline E } \cdot d\overline l

[/tex]

where C is a reference point, which I set to zero, which is at one of the plates.

I get that 500 V = -E*d. Why do I get a minus, when it should be positive (I know that it must be a positive number).

3) Is it possible to have negative capacitance? I would personally think no, when thinking of the physical meaning of capacitance, but I need to be sure.

I have some questions about electrostatics, and I hope you can help me. Here they are:

1) Please take a look at the following example: http://books.google.dk/books?id=YkGQFFhZWicC&pg=PA387&lpg=PA387&dq="electric+field+produced+by+a+uniformly+polarized"&source=web&ots=4pobXLVw66&sig=YiULcbQhTa7htSGbHJ4-D4ZBaJU&hl=da

Here we are calculating the electric field

**E**of a uniformly, polarized solid sphere. My question is: When it is uniformly polarized, there are**no**volume bound charges. Then how can we have an electric field inside the sphere? Isn't the setup equivalent of a spherical shell, which has E=0?2) I have two conducting metal plates with a distance

*d*between them. I put a potential of 500 V on the plates, and I want to find the electric field. I use the following[tex]

V = - \int_C^r {\overline E } \cdot d\overline l

[/tex]

where C is a reference point, which I set to zero, which is at one of the plates.

I get that 500 V = -E*d. Why do I get a minus, when it should be positive (I know that it must be a positive number).

3) Is it possible to have negative capacitance? I would personally think no, when thinking of the physical meaning of capacitance, but I need to be sure.