Derive the expression for the electric field

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The discussion focuses on deriving the electric field inside a long, solid cylinder with uniform positive charge density ρ. Using Gauss's law, a cylindrical Gaussian surface is chosen, where the electric displacement field D is constant and perpendicular to the surface. The total charge enclosed by the Gaussian surface is calculated as Q = ρ * V, with V being the volume of the cylinder segment. The area of the Gaussian surface is approximated by A = 2 * π * r * h, leading to the expression D = Q/A = (ρ * r) / 2. Finally, the electric field E is derived as E = (ρ * r) / (2 * ε₀).
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Homework Statement



A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rho

Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rho.



The Attempt at a Solution


\rho/(r*2*pi*\epsilon_0)
 
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Do you know Gauss's law? For an appropriate gaussian surface (D-field perpendicular to the guassian surface and constant everywhere on the surface) DA = Q or D = Q/A, where Q is the total charge enclosed by the Gaussian surface and A is the surface area of the Gaussian surface.

Choose a cylindrical Gaussian surface and assume that a negligible amount of flux passes through the end caps, since it is a 'very long' cylinder (meaning the end cap area is small compared to the total area of the cylinder).

Q is the charge density multiplied by the volume of the Gaussian surface. V = h*pi*r^2, where 'r' is the distance from the axis of the cylinder. So Q = rho*h*pi*r^2.

A is the area of the gaussian surface (neglecting endcap area). A = 2*pi*r*h.

So, D = Q/A = (rho*r)/2

The relationship between D and E is E = D/ep_0, so E = (rho*r)/(2*ep_0)
 
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