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Topic: Off Topic Physics/math question
Replies: 3   Last Post: Jan 5, 2013 12:35 PM

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Scott Berg

Posts: 2,111
Registered: 12/12/04
Re: Off Topic Physics/math question
Posted: Jan 5, 2013 12:28 PM
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<> wrote in message
> ---
> 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.

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