electric potential

by Miraj Kayastha
Tags: direction, electric, electric potential, field, potential, strength
 P: 54 is electric field strength always equal to negative potential gradient or can it be equal to positive potential gradient sometimes?
 PF Gold P: 369 The electric field is a vector quantity, the negative sign shows that the direction of E is opposite to the direction in which V increases. This relationship is true in any circumstance as far as I know there isn't any exception.
 P: 54 Isn't the formula for finding the electric field strength between two parallel plates E =V/d
PF Gold
P: 369

electric potential

It is an equation for the magnitude. The electric field is a vector the voltage is a scalar. You have to assign the direction. When discussing the E field as a vector you would attach a negative sign actually.

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
 P: 54 why is there a negative sign in the equation E = -V/d?
 Mentor P: 11,255 The electric field E has both magnitude and direction. Its direction is the direction in which the electric potential V decreases.
 P: 998 E = -grad V holds for all conservative E fileds, that is E fields due to charged particles, ions, etc. But for the case of induction involving time varying magnetic fields, E is not the gradient of potential V. E is the negative partial derivative of A, the magnetic vector potential wrt time. If we have 2 coils separated by air, the primary is excited by a power source, let's say constant voltage supply, ac, the secondary coil has an E field present, as well as a current and voltage due to induction. This E field is not the gradient of the scalar electric potential V. It is as follows: E = -dA/dt. Any e/m fields text will elaborate. Claude
 Mentor P: 16,489 For completeness ##\mathbf{E}=-\nabla V - \partial \mathbf{A}/\partial t## 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.
P: 998
 Quote by DaleSpam For completeness ##\mathbf{E}=-\nabla V - \partial \mathbf{A}/\partial t## 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:
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
Mentor
P: 28,841
 Quote by Miraj Kayastha is electric field strength always equal to negative potential gradient or can it be equal to positive potential gradient sometimes?
This is a very strange question. It is as if you are asking if this description is negotiable.

Zz.
 P: 10 It is a sign convention. Remember that $V=- \int E.dl$ 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
 Sci Advisor PF Gold P: 11,398 That "sign convention" is based in the definition, which is the work done ON a unit positive charge. That's the only arbitrary bit. All the rest follows.

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