Deriving Electric Field for a Conductor-Shell System using Gauss's Law

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The discussion focuses on deriving the electric field for a conductor-shell system using Gauss's Law. A long straight conducting rod with a linear charge density of +2.0 µC/m is enclosed by a thin cylindrical shell with a linear charge density of -2.0 µC/m. The electric field inside the shell is calculated using the formula E = q/(4πε₀r²), resulting in E = 4.5 x 10⁹ N·m²/C. The application of Gauss's Law, specifically the integral form ∮ E·dA = Q_encl/ε₀, is emphasized for determining the electric field both inside and outside the cylindrical shell.

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A long straight conducting rod (or wire) carries a linear charge density of +2.0uC/m. This rod is totally enclosed within a thin cylindrical shell of radius R, which carries a linear charge density of -2.0uC/m.
A) Construct a Gaussian cylindrical surface between the rod and the shell to derive then electric field in the inner space as a function of the distance from the center of the rod.
B) Construct a Gaussian cylindrical surface outside both the rod and the shell to calculate the electric field outside the shell.

This is what i have so far.

E=q/4piEor^2
E=+2.0uC/m / 4pi8.85x10^-12(-2uC/m)^2
E=4.5x10^9Nm^2/C
 
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chillaxin said:
A long straight conducting rod (or wire) carries a linear charge density of +2.0uC/m. This rod is totally enclosed within a thin cylindrical shell of radius R, which carries a linear charge density of -2.0uC/m.
A) Construct a Gaussian cylindrical surface between the rod and the shell to derive then electric field in the inner space as a function of the distance from the center of the rod.
B) Construct a Gaussian cylindrical surface outside both the rod and the shell to calculate the electric field outside the shell.

This is what i have so far.

E=q/4piEor^2
E=+2.0uC/m / 4pi8.85x10^-12(-2uC/m)^2
E=4.5x10^9Nm^2/C
It asks you to apply Gauss' law:

\oint E\cdot dA = \frac{Q_{encl}}{\epsilon_0}

If you pick a gaussian surface through which you know E is constant due to all points on the surface being equidistant from equal charges, the integral is simply

E\cdot A = \frac{Q_{encl}}{\epsilon_0}

Pick a surface that is inside the cylinder that fits that description and do the calculation.

AM
 
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