Derive the expression for the electric field

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SUMMARY

The electric field inside a long, solid cylinder with a uniform positive charge density \(\rho\) can be derived using Gauss's law. By selecting a cylindrical Gaussian surface, the total charge \(Q\) enclosed is calculated as \(Q = \rho \cdot h \cdot \pi r^2\), where \(h\) is the height of the cylinder and \(r\) is the distance from the axis. The area \(A\) of the Gaussian surface is \(A = 2 \pi r h\). Consequently, the electric displacement field \(D\) is expressed as \(D = \frac{Q}{A} = \frac{\rho r}{2}\), leading to the final electric field \(E = \frac{D}{\epsilon_0} = \frac{\rho r}{2 \epsilon_0}\).

PREREQUISITES
  • Understanding of Gauss's law in electrostatics
  • Familiarity with electric displacement field (D-field)
  • Knowledge of charge density (\(\rho\)) and its implications
  • Basic concepts of cylindrical coordinates in physics
NEXT STEPS
  • Study the application of Gauss's law in different geometries
  • Explore the relationship between electric displacement field and electric field
  • Investigate the effects of varying charge densities on electric fields
  • Learn about boundary conditions in electrostatics
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Physics students, electrical engineers, and anyone studying electrostatics and electric field calculations in cylindrical geometries.

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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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