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- Thread starter Miraj Kayastha
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FOIWATER

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Isn't the formula for finding the electric field strength between two parallel plates E =V/d

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FOIWATER

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In fact you need it in that case. The E field increases towards lower potential (means a positive direction is pointing in the direction of E, that is, to a lower potential). In V=-Ed, a negative direction (corresponding to moving towards a positive potential) will give you a positive result for V. if you are moving away from the higher potential, a positive direction, the potential is lower. So I believe it is just a matter of calculating magnitudes vs. considering it as a vector.

If I have misled you someone will clarify it for us both. I am no expert but I just did an electromagnetics class

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why is there a negative sign in the equation E = -V/d?

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jtbell

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This E field is not the gradient of the scalar electric potential V. It is as follows:

Any e/m fields text will elaborate.

Claude

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So the sign of ##\nabla V## is always negative, never positive, but there is an additional term besides just the negative gradient of the potential. This is what cabraham is talking about.

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So the sign of ##\nabla V## is always negative, never positive, but there is an additional term besides just the negative gradient of the potential. This is what cabraham is talking about.

Yes of course. The 1st term:

-grad V

refers to the conservative portion of the total E field. The 2nd term:

-dA/dt

refers to the non-conservative portion of said E field. An example would be where a loop is immersed into a time changing magnetic field. The loop is circular, where each half of the loop is a different resistivity material semicircular in shape. Of course the 2 semicircular loops are in series so they have the same current, but the voltage across each half must be equal as they are also in parallel. This condition is met by virtue of charge layer accumulation at the 2 boundaries between the media. The electrons moving around the loop feel the Lorentz force from both the time varying B field as well as the static charges accumulated at the boundaries.

The result is that there are 2 E fields, one from induction, non-conservative, given by the 1st term above, and a 2nd, conservative, per 2nd term above. Interesting stuff this is.

Claude

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This is a very strange question. It is as if you are asking if this description is negotiable.

Zz.

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It is a sign convention. Remember that [itex] V=- \int E.dl [/itex] the minus sign is place there so that when we bring a charge from infinity (where the potential is defined as 0) and place it at some point in space we arrive at some positive value for V and therefore the work we have done is positive

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sophiecentaur

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