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Question

(B) The electric field intensity $E$ at a perpendicular distance $r$ from an infinitely long,straight,uniformly charged wire with linear charge densit

Vedclass · Accepted Answer

$\frac{1}{r}$

Vedclass · Answer

(B) The electric field intensity $E$ at a perpendicular distance $r$ from an infinitely long,straight,uniformly charged wire with linear charge density $\lambda$ is given by Gauss's Law as:$E = \frac{\lambda}{2 \pi \varepsilon_{0} r}$Here,$\lambda$ is the linear charge density,$\varepsilon_{0}$ is the permittivity of free space,and $\pi$ is a constant.Since $\lambda$,$\pi$,and $\varepsilon_{0}$ are constants,the relationship between the electric field and distance is:$E \propto \frac{1}{r}$Therefore,the electric field varies inversely with the distance $r$.

Electric field due to an infinite,straight,uniformly charged wire varies with distance $r$ as:

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