I tried it with no success..
I attempted to treat it like a ring of charge...finding E then integrating to find V.
Still no success.
I think I am getting lost in the "point any where on the axis inside or outside the tube".
any suggestions how to solve this?
Allow me to expand on what genneth started..
he is correct...E= - Del V
where del is the gradiate operator.
You have V in cartesian.
so del V = x(hat)*dV/dx+y(hat)*dV/dy+z(hat)*dV/dz
You text book should give the relationship for del V, don't forget the negative sign.
So take the...
Here is the question. a hollow, thin walled insulating cylinder of radius b and height h has charge Q uniformly distributed over its surface. Calculate the electric potential and field at all points along the z axis of the tube.
Outside the tube
Inside the tube.
I know how to find the...
Sort of
I think I see what is going on...I have to take the partial derivative of each value with respect to the cooresponding value, r to x theta to x and phi to x, then r, theta, and phi to y and then to z...this in turn will provide the conversion, right?
Is this essentially what how you...
Do I have to do this for y and z?
dT/dy= dT/dR*dR/dy+dT/d(theta)*d(theta)/dy+dT/d(phi)*d(phi)/dy
dT/dz= dT/dR*dR/dz+dT/d(theta)*d(theta)/dz+dT/d(phi)*d(phi)/dz
Then after this, its gets ugly, but I think this is how you do it...
x^=R^(cos...)+theta^(cos...)-phi^(sin phi)...
Correct so far?
First, thanks for responding...
Second, am I correct so far?
Third, I don't quite understand what you mean, could you expand on that?
Is it these values substitute into right hand side?
x=R sin theta cos phi
y=R sin theta sin phi
z=R cos theta
I am trying, but I...
I need to derive the expression for the gradient operator in spherical coordinates.
I know the following
R =sqrt(x^2+y^2+z^2)
theta, call it %, = arctan sqrt(x^2+y^2)/z
phi, arctan (y/x)
Using dT/dx= dT/dR*dr/dx+dT/d%*d%/dx+dT/dphi*dphi/dx, do partial derivates...
dR/dx =...
Got it...
I was making it harder than it was...
I solve for one of the variables, say a....
a=b+2c-d
Then I plug this value into L.
[b+2c-d, b,c,d]
I could set up in aug matrix, but I can pull the constants b, c, d right out to get the kernel basis.
Then my vectors are...
I think I have it...if there is constants for artibary numbers then those constants make up the kernal and the other is the range, correct?
SO if the the dim is 4 and the kernel is three, then the range is one...
if I solve it for a=b+2c-d, then the kernal is the constants, b, c, d times...
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I know the dimension of the kernel is 3, but how?
I have tried setting it...
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Let T ={ [1, 0], [1, 1] }be a basis for R2 .
Given that Transition matrix P s←t
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