A long charged cylinder of linear charge dens | vedclass

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(A) Consider a Gaussian surface in the form of a cylinder of radius $r$ and length $L$ coaxial with the charged cylinder,where $r$ is the distance bet

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$E = \frac{\lambda}{2 \pi \epsilon_{0} r}$

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(A) Consider a Gaussian surface in the form of a cylinder of radius $r$ and length $L$ coaxial with the charged cylinder,where $r$ is the distance between the two cylinders.According to Gauss's Law,the total electric flux $\phi$ through the Gaussian surface is given by $\phi = \oint E \cdot dA = E(2 \pi r L)$.The charge enclosed by this Gaussian surface is $q_{enclosed} = \lambda L$.Applying Gauss's Law,$\phi = \frac{q_{enclosed}}{\epsilon_{0}}$,we get $E(2 \pi r L) = \frac{\lambda L}{\epsilon_{0}}$.Solving for $E$,we find $E = \frac{\lambda}{2 \pi \epsilon_{0} r}$.Thus,the electric field in the space between the two cylinders is $\frac{\lambda}{2 \pi \epsilon_{0} r}$.

$A$ long charged cylinder of linear charge density $\lambda$ is surrounded by a hollow coaxial conducting cylinder. What is the electric field in the space between the two cylinders?

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