- #1
Ozmahn
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Homework Statement
A straight, nonconducting plastic wire 9.50cm long carries a charge density of 100 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.
Find the magnitude and direction of the electric field this wire produces at a point 6.00cm directly above its midpoint.
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 6.00cm directly above its center.
##\lambda=100 {\frac{nC}{m}}##
Homework Equations
Area of cylinder
## A=\pi r^2*h##
##\oint \vec E \cdot d \vec A = {\frac{Q_{enclosed}}{\epsilon_0}}##
The Attempt at a Solution
Create a cylinder around wire, with end caps at either end of the wire.
##\oint \vec E \cdot d \vec A = {\frac{Q_{enclosed}}{\epsilon_0}}##
since E is perpendicular to dA
E is constant at the radius of P (.06m), so you can factor it out of the integral.
##\ E \oint \cdot dA##
Integrating dA gives you A. Area of a cylinder is given by
## A= \pi r^2*h##
##E* \pi r^2*h={\frac{Q_{enclosed}}{\epsilon_0}}##
Since Q is the same as the charge density (##lambda##) times length,
##Q=\lambda*0.095##
The height of the cylinder is equivalent to the length of the wire, so
##h=l=0.095##
Sub that into Gauss' law, and I have
##E= {\frac{\lambda*l}{\pi*r^2*\epsilon_0 (l)}} ##
The lengths (0.095) cancel, and I separate the k constant from the rest
##E= {\frac{\lambda}{r^2}} * {\frac{1}{\pi \epsilon_0}}##
##{\frac{1}{\pi \epsilon_0}}=4*{\frac{1}{4\pi \epsilon_0}} = 4k = 4*8.99*10^9##
Plugging this in, my equation looks like
##{\frac{100*10^{-9}*(4*8.99*10^9)}{0.06^2}} = 998888.8889 ##
That's not the right answer, but I'm not sure where I messed up. Any help is appreciated, thanks.