Calculate the electric field strength inside and outside a wire

AI Thread Summary
To calculate the electric field strength inside and outside an infinitely long conducting cylinder with a uniform surface charge density (lambda), it's important to recognize that the electric field inside the conductor is zero due to electrostatic shielding. For the region outside the cylinder, the electric field can be determined using Gauss's law, where the charge enclosed by a Gaussian surface is calculated based on the line charge density. The correct expression for the electric field outside the cylinder is E = (lambda)/(2*pi*epsilon_0*r), where r is the distance from the center of the cylinder. The discussion highlights the need to properly define the Gaussian surface and the area when applying Gauss's law to find the electric field. Understanding these concepts is crucial for solving problems related to electric fields around charged conductors.
matt_crouch
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Homework Statement



An infinitly long conducting cylinder of Radius R carries a inform surface charge of (lambda per unit length) determine the electric field strength inside and outside the cylinder

Homework Equations



integral (E.ds)= q/e0

The Attempt at a Solution



im not really sure what to do at all i tried simply differentiating the equation above and substituting lambda in for q but I am pretty sure that's not right...
 
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First, the field inside is easiest to start with. The q in integral (E.ds) = q/e0, is the charge inside a gaussian surface. So a gaussian surface inside the conductor encloses how much charge? ... so the electric field is...?

Next, you need to simply integrate the left side of the equation after subtituting dr for ds. Then use a gaussian surface of some length l, and figure out what the q enclosed is and plug that into the right side.
 
so since its a Gaussian surface do i calculate the line charge density by integrating lambda from R to infinity? then substitute that into the equation above?
 
Electric field inside the cylinder is zero. Because that is conductor material
 
ok so outside the conductor do i just use the E=q/4(pi)e0R^2 ?
 
matt_crouch said:
so outside the conductor do i just use the E=q/4(pi)e0R^2 ?
It's not correct. What is area of Gaussian surface?
 
ahhh so i need to use the area of a cylinder? so 2(pi)rh +2(pi)r^2 ?
 
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