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Re: The sign of work



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In trying to justify the negative sign in dV=-E*ds, where
E and ds are the electric field and the displacement, I will
emphasize the dot product nature of E*ds. The potential
energy (per unit charge) increases when the dot product
is negative (obtuse angle between E and ds), otherwise
it decreases, or remains the same. The "uphill" is locally
defined by the direction of E; it has nothing to do with
the orientation of the x,y,z axes (with respect to local E).
There is no ambiguity here, as long as we agree that the
probe charge is positive. Right?



There is no ambiguity, but the reason for the min8us sign, really is the
fact that potential is defined to be zero at infinite distance from the
charge.
James k.