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(B) For an infinitely long uniformly charged wire with linear charge density $\lambda$,we use Gauss's Law to find the electric field at a distance $r$

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$E \propto \frac{1}{r}$

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(B) For an infinitely long uniformly charged wire with linear charge density $\lambda$,we use Gauss's Law to find the electric field at a distance $r$.Consider a cylindrical Gaussian surface of radius $r$ and length $l$ coaxial with the wire.The total electric flux $\Phi_E$ through the Gaussian surface is given by $\Phi_E = E 	imes (2\pi rl)$.According to Gauss's Law,$\Phi_E = \frac{q_{enclosed}}{\epsilon_0}$.Since the charge enclosed is $q_{enclosed} = \lambda l$,we have $E 	imes (2\pi rl) = \frac{\lambda l}{\epsilon_0}$.Solving for $E$,we get $E = \frac{\lambda}{2\pi \epsilon_0 r}$.Therefore,the electric field $E$ is inversely proportional to the distance $r$,i.e.,$E \propto \frac{1}{r}$.

The electrostatic field of a long uniformly charged wire varies with distance $r$ according to which relation?

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