The electric field intensity at a point at a | vedclass

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(C) For an infinitely long charged wire or pipe with linear charge density $\lambda$ (given as $q$ per unit length),we use Gauss's Law.Consider a cyli

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Inversely proportional to $r$

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(C) For an infinitely long charged wire or pipe with linear charge density $\lambda$ (given as $q$ per unit length),we use Gauss's Law.Consider a cylindrical Gaussian surface of radius $r$ and length $L$ coaxial with the pipe.The total electric flux $\Phi_E$ through the Gaussian surface is given by $\Phi_E = E \cdot (2\pi r L)$.According to Gauss's Law,$\Phi_E = \frac{q_{enclosed}}{\epsilon_0} = \frac{\lambda L}{\epsilon_0}$.Equating the two expressions: $E(2\pi r L) = \frac{\lambda L}{\epsilon_0}$.Solving for $E$,we get $E = \frac{\lambda}{2\pi \epsilon_0 r}$.Therefore,$E \propto \frac{1}{r}$,which means the electric field is inversely proportional to the distance $r$.

The electric field intensity at a point at a distance $r$ from the axis of an infinitely long pipe having a linear charge density $q$ is proportional to:

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