Charged Cylinder

Electric Field

Topic reserved by Jennifer Burkhardt

The electric field of a uniformly charged cylinder of length [math]\displaystyle{ L }[/math] and radius [math]\displaystyle{ R }[/math] can be found through viewing the cylinder as a collection of long uniformly charged rods forming a circle or a series of uniformly charged rings stacked on top of another. Before determining which method to use, the point of observation must be decided.

Uniformly Charged Rods

If measuring the electric field of a cylinder from an observation point that is at the center of the cylinder, the method of rods would be most useful.

L≫R

The location vector from the center of a rod can be determined in terms of the angle θ.

[math]\displaystyle{ \vec{r} = \lt 0,0,0\gt - \lt 0,R*sinθ,R*cosθ\gt = \lt 0,-R*sinθ,-R*cosθ\gt }[/math] [math]\displaystyle{ \hat{r} = \frac{\vec{r}}{r} = \lt 0,-sinθ,-cosθ\gt }[/math]

The amount of charge on one of the rods, assuming that the cylinder is complete, is given by:

[math]\displaystyle{ dQ = Q\frac{dθ}{θ(total)} = \frac{Qdθ}{θ(total)} }[/math]

[math]\displaystyle{ dθ }[/math] is the angular width of one rod, and [math]\displaystyle{ θ(total) }[/math] is the angular extent of the cylinder.

The contribution of this individual rod to the total electric field,, assuming that the cylinder is complete, is:

[math]\displaystyle{ \Delta \vec{E} = |\Delta \vec{E}| \vec{r} ≈ \frac{1}{4π\epsilon_0} \frac{2dQ}{L} \frac{1}{R} \lt 0,-sinθ,-cosθ\gt = \frac{1}{4π\epsilon_0} \frac{2}{LR} \frac{Qdθ}{θ(total)} \lt 0,-sinθ,-cosθ\gt }[/math]

To find the total electric field from all of the rods, the integral of the electric field will have to be taken from the x, y, and z components separately.

[math]\displaystyle{ E_x = \int\limits_0^{θ(total)}\ \frac{1}{4π\epsilon_0} \frac{2}{LR} \frac{Q}{θ(total)} * 0dθ = 0 }[/math]

[math]\displaystyle{ E_y = \int\limits_{0}^{θ(total)}\ \frac{1}{4π\epsilon_0} \frac{2}{LR} \frac{Q}{θ(total)} * (-sinθ)dθ = \frac{1}{4π\epsilon_0} \frac{Q}{LRπ} cosθ|_{0}^{θ(total)} }[/math]

[math]\displaystyle{ E_z = \int\limits_{0}^{θ(total)}\ \frac{1}{4π\epsilon_0} \frac{2}{LR} \frac{Q}{θ(total)} * (-cosθ)dθ = \frac{-1}{4π\epsilon_0} \frac{Q}{LRπ} sinθ|_{0}^{θ(total)} }[/math]

Without Approximation

The location vector from the center of a rod can be determined in terms of the angle θ.

[math]\displaystyle{ \vec{r} = \lt 0,0,0\gt - \lt 0,R*sinθ,R*cosθ\gt = \lt 0,-R*sinθ,-R*cosθ\gt }[/math] [math]\displaystyle{ \hat{r} = \frac{\vec{r}}{r} = \lt 0,-sinθ,-cosθ\gt }[/math]

The amount of charge on one of the rods, assuming that the cylinder is complete, is given by:

[math]\displaystyle{ dQ = Q\frac{dθ}{θ(total)} = \frac{Qdθ}{θ(total)} }[/math]

[math]\displaystyle{ dθ }[/math] is the angular width of one rod, and [math]\displaystyle{ θ(total) }[/math] is the angular extent of the cylinder.

The contribution of this individual rod to the total electric field,, assuming that the cylinder is complete, is:

[math]\displaystyle{ \Delta \vec{E} = |\Delta \vec{E}| \vec{r} = \frac{1}{4π\epsilon_0}\frac{dQ}{R*sqrt(R^2+(L/2)^2)} \lt 0,-sinθ,-cosθ\gt = \frac{1}{4π\epsilon_0} \frac{1}{R*sqrt(R^2+(L/2)^2)} \frac{Qdθ}{θ(total)} \lt 0,-sinθ,-cosθ\gt }[/math]

To find the total electric field from all of the rods, the integral of the electric field will have to be taken from the x, y, and z components separately.

[math]\displaystyle{ E_x = \int\limits_0^{θ(total)}\ \frac{1}{4π\epsilon_0} \frac{1}{R*sqrt(R^2+(L/2)^2)} \frac{Q}{θ(total)} * 0dθ = 0 }[/math]

[math]\displaystyle{ E_y = \int\limits_{0}^{θ(total)}\ \frac{1}{4π\epsilon_0} \frac{1}{R*sqrt(R^2+(L/2)^2)} \frac{Q}{θ(total)} * (-sinθ)dθ = \frac{1}{4π\epsilon_0} \frac{Q}{LRπ} cosθ|_{0}^{θ(total)} }[/math]

[math]\displaystyle{ E_z = \int\limits_{0}^{θ(total)}\ \frac{1}{4π\epsilon_0} \frac{1}{R*sqrt(R^2+(L/2)^2)} \frac{Q}{θ(total)} * (-cosθ)dθ = \frac{-1}{4π\epsilon_0} \frac{Q}{LRπ} sinθ|_{0}^{θ(total)} }[/math]

Uniformly Charged Rings

If measuring the electric field of a cylinder from an observation point that is away from the cylinder in the z-direction (assuming that the circles the cylinder is composed of reside in the x-y plane), the method of rings will be most useful.

Without Approximation

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Examples

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Simple

Find the net electric field at the center of a half cylinder, projected on the z-axis, that is 2 meters long, has a radius of .05 meters (2≫.05), and has a 3 Coulomb charge.

Solution

[math]\displaystyle{ dQ = Q\frac{dθ}{π} = \frac{Qdθ}{π} }[/math]

[math]\displaystyle{ \vec{E}_{net} = \lt x,y,z\gt }[/math]

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