The electric intensity due to an infinite cyl | vedclass

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Question

(c) According to Gauss law $\oint {E\, \cdot \,ds} $ $==>$ $ \frac{{ql}}{{{\varepsilon _0}}}$
$\oint {ds = 2\pi \,rl;} $ ($E$ is constant)

Vedclass · Accepted Answer

Inversely proportional to $r$

Vedclass · Answer

(c) According to Gauss law $\oint {E\, \cdot \,ds} $ $==>$ $ \frac{{ql}}{{{\varepsilon _0}}}$
$\oint {ds = 2\pi \,rl;} $ ($E$ is constant)
 $E \cdot \,2\pi \,rl = \frac{{ql}}{{{\varepsilon _0}}}$ $ ==> $ $E = \frac{q}{{2\pi {\varepsilon _0}r}}$ i.e. $E \propto \frac{1}{r}$

The electric intensity due to an infinite cylinder of radius $R$ and having charge $q$ per unit length at a distance $r(r > R)$ from its axis is

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