How Does Charge Density Affect Electric Field Calculations?

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Homework Statement



A straight, nonconducting plastic wire 8.00 cm long carries a charge density of 125 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.

A) Find the magnitude and direction of the electric field this wire produces at a point 5.50 cm directly above its midpoint.

B) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 5.50 cm directly above its center.

I have tried to figure it out, but I am having some issues.

Homework Equations



theta = r/(L/s)

E = Q/2πrεoL = ρL/2πrL = ρ/2πεor

The Attempt at a Solution



E = 125*10^-9/((2pi(0.055)8.85*10^-12 = 4.0871*10^4

Ey = Esin(theta)

theta = r/(L/s) = arctan(2*0.055)/0.08 =arctan(1.375) = 53.97 degrees

Ey= 4.0871*10^4 * sin53.97 degrees = 3.31*10^4

Alternatively, would I do arctan again for 53.57 degrees for 88.938 degrees? Then multiply it by 4.0871*10^4? The answer would be 4.086*10^4

Is this right?? Please help me.

b)

r= L/2pi = 0.08/2pi = 0.0126 m

d = square root(0.0126^2+0.055^2) = 0.0564 m

Ey = Esin(theta)

theta = arctan(0.055)/(0.0126m) = 77.1 degrees

Ey = PL/4piEor^2 = 125*10^-9 * 0.08 / 4pi(8.85*10^-12(0.055^2) = 2.97*10^4
 
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It is usually easier to think about to put your rod in an explicit coordinate system - I'd suggest along the z axis of cylindrical-polar coordinates with the com of the rod at z=0.

So the rod goes from -4 to 4 cm, and you want the field at point P=(r,θ,z)=(5.5,0,0).

You need to sum the infinitesimal contributions across the length of the rod - like this:[tex]E=\int_{z=-4}^{z=4}d\vec{E}[/tex]
Where dE is contribution to the field at P due to an arbitrary bit of the rod at position Z=(0,0,z), length dz. At that position, for example, Q(z)=ρdz where ρ is the linear charge density. The square distance will be |ZP|2 = (z2+5.52) ... from which you get the magnitude.

Since this will be a vector sum, you need to resolve into components - pick radial and z-axis components. You'll find the sum of the z-axis components will be zero.

I think where you are going wrong, if I read you right, is that you have fixed your theta when it will actually vary with z. Express the sine as a function of z and you'll get there. I've just used theta as a coordinate label so I have to change notation:

If [itex]\alpha[/itex] is the angle between [itex]\vec{ZR}[/itex] and [itex]\vec{OZ}[/itex] [recall O=(0,0,0)] then [itex]\int dE\sin\alpha = 0[/itex] and [itex]\cos\alpha[/itex] is determined from trigonometry.

Note: it will help you understand if you draw a picture.
 
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Alright, thank you very much for the detailed and well-written response. I have a much better idea on what is actually happening.