<email@example.com> wrote in message news:firstname.lastname@example.org... > --- > If y(x) = 1 / x > then y(x) > 0 when x > 0 and y(x) < 0 when x < 0. > > According to the power rule of Calculus, y'(x) = -1 / x^2. > So y'(x) is always negative. > > Are we losing information by taking the derivative?
what do you mean, "losing information" ? If you take the integral of a derivative you need to add a constant C for a particular case, everyone alrady knows that.
> Here is where the physics comes in. > > Let E(x, y, z) be "electric electric field" (a vector). > > Let Ex be the x component of E, etc. We will have a charge of "q" > at the origin and compute the electric field at an arbitrary point > due to it. > > Then > Ex = k q x / |x|^3
what dose that mean ? Esubx = k * q * (x /( |x|^3)) ??
> Ey = k q y / |x|^3 > Ez = k q z / |x|^3 > > I am told that there is this thing called electric potential energy, > but I am having mathematical trouble figuring out how it relates to > the electric field E.
this is an electrical engineering question
> > Let electric potential p(x,y,z) be defined such that > E = -grad p
grad p ?? do you mean grad( p ) ??
> > It looks to me like we're in trouble now. > For according to > Ex = k q x / |x|^3 > > Ex is positive when x > 0 and negative when Ex is < 0. This makes > sense, since a positive charge to the right of our origin (where we > have placed a charge of +q) will want to go right, in the +x direction, > whereas a detector charge placed to the left of our origin will want to > go left, i.e. Ex < 0 if x < 0 and Ex > 0 if x > 0. > > I'm told the formula for p is > p = k q / r
p = (k * q) / r ??
> > We should then have E = -grad p. > > I think the place I make a mistake is in my understanding of how r relates > to x. > If everything is oriented along the x axis, aren't they the same? Or do we > have > r = |x| ?
no, r is vector in x,y,z
> > If we have r = x we run into trouble: > > p = k q / x > grad p = (-k q) / x^2 > So -gradp = kq / x^2 > > This is supposed to be the same as Ex but we have > Ex = k q x / |x|^3 > > So indeed they are the same if x > 0. > > Obviously I need to study my calculus and physics, but can anyone shed > some light into this? > > My concern is that if r = x because we have both charges on the x axis > then I lose information; > when x < 0, Ex is < 0 by definition but starting with p = k q / x I find > dp/dx is positive! > > So my assumption that r = x is clearly wrong. > > Maybe r = |x|. > > If r = |x|, then I still don't see how I can recover E = (Ex, Ey, Ez) > from -grad p. > > My understanding is that the electric potential scalar and electric field > vector can be > converted from one to the other. > > So my real question is: am I to understand the electric potential due to a > positively charged > point charge is positive everywhere, because we use r = sqrt(x^2 + y^2 + > z^2) in the formula and > not e.g. r = x when it's in the x axis? > > If so, suppose I give you ONE point and don't tell you where it is. This > is the point at which > I take an electric potential scalar reading. The point charge producing > the potential is at the > origin and you know what it's charge is. Now all I tell you is the scalar > potential. > > Am I to understand then that from this scalar you CANNOT tell me the > electric field vector? I.e. > you cannot convert one to another with quantities, but you'd need another > nearby point's potential > reading to do so? > > It looks to me like if I tell you the electric potential scalar value is A > that you will have to say: > well I know |Ex| but have no idea about Ex: it could be +|Ex| or -|Ex| > ???? > > Is this really the way the math works out, or am I missing something? > I just kind of assumed, that E = -grad p meant that given p you could find > E. > I guess instead what is meant, is: given all of p you can find all of E.