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# Coefficients of the solution to a second order differential equation watch

1. Ok I have this function now which is given as a function of the position vector r and it says

phi(r)=((A/r^2)+Br)[email protected]

phi(0)=0

dphi(0)/dz=1
2. huh ... phi(r) = (A/r^2 ...)
phi(0) = (A/0^2 ...) ---> not defined?
3. Well phi(0)=0 provided A=0, so you have phi(r) = Brcos(t).

Not sure about the d/dz part, considering its a function of t (theta).
4. (Original post by AlphaNumeric)
Well phi(0)=0 provided A=0, so you have phi(r) = Brcos(t).

Not sure about the d/dz part, considering its a function of t (theta).
That's what I was thinking of... but then don't know what z is here.
5. r^2 = x^2 + y^2 + z^2 right?
6. Could be, but then if you're working in polar or spherical polar coords, it's terribly bad notation to take a derivative with respect to a cartesian coordinate. Why not define the boundary condition another way? Seems like it's the only logical approach though, so the question seems a bad one.
7. I will post the full question:

In electrostatics, the electric potential field phi(r) satisfies the equation (del^2)(phi) = 0 in every region of space where there are no charges. In a certain charge-free region, the electrostatic potential is given by phi(r) = f(r) [email protected], where r is the distance from the origin and @ is the polar angle of spherical polar coordinates. Starting from the Laplacian operator del^2 in spherical polar coordinates, find the second-order ordinary differential equation satisfied by f(r). Hence, determine the function f(r), with any arbitrary constants fixed by the conditions that phi = 0 and dphi=dz = 1 V/m at the origin.

We have both gotten as far as working out the ODE: (r^2)f''(r)+2rf'(r)-2f(r)=0

Then it gets tricky.

8. (Original post by Νεωτον)
Ok I have this function now which is given as a function of the position vector r and it says

phi(r)=((A/r^2)+Br)[email protected]

phi(0)=0

dphi(0)/dz=1
My conclusion is as follows

phi(0)=((A/r^2)+Br)[email protected]=0

There exists no such solution unless A=0

Hence phi(r)[email protected]

dphi(r)/dz=d(B(x+y+z)[email protected])/[email protected]

dphi(0)/dz=Bcos0=B*1=1=>B=1

t.f. f(r)=r
9. Assuming partial derivatives, dr/dz = z/r, not 1.

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Updated: February 27, 2006
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