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chipotleaway
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Homework Statement
Find the electric field at a point on the x-axis due to a uniformly charged, hollow cylinder. The x-axis is parallel to the cylinder and goes through it's centre. One end of the cylinder is located at the origin.
Homework Equations
R = radius of cylinder
Q = charge on cylinder
L = length of cylinder from the origin
D = distance from the origin to the point of interest
k = 1/4πε
The Attempt at a Solution
1) I found the charge density
[itex]\lambda=\frac{Q}{2\pi RL[/itex]
2) So an infinitesimally thin slice of the cylinder 2πRdx would have charge
[itex]dQ=\frac{Q}{L}dx[/itex]
3) The force due to one of these rings is
[itex]dF = k\frac{qdQ}{(D-L)^2+R^2}[/itex]
4) We want only the horizontal component due to the symmetry of the cylinder and the cancellation of the vertical components of the forces. So since
[itex]cos(\theta) = \frac{D-L}{((D-L)^2+R^2)^{\frac{1}{2}}[/itex]
Therefore [itex]dF_x=k\frac{qdQ(D-L)}{((D-L)^2+R^2)^{\frac{3}{2}}[/itex]
I divided by q to get the electric field, replaced dQ with (Q/L)dx and then integrated with respect to x from 0 and L.
The formula I got is:
[itex]E=k\frac{Q(D-L)}{((D-L)^2+R^2)^\frac{3}{2}[/itex] which doesn't seem right or 'behave' right. Based on what I know, if we used it to calculate the force on a test charge and we let the test charge be a very large distance away, then we should be left with Coulomb's law but that doesn't seem to happen with what I got.