The electric field intensity due to an infinite rod of linear charge density $γ$ at a distance $r$ is $E$ . Then the electric field intensity due to a semi-infinite rod of linear charge density $γ$ at point $P$ , at a perpendicular distance $r$ is

\frac{E}{2}

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\frac{E}{\sqrt{2}}

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E

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\frac{E}{2 \sqrt{2}}

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Solution

The correct option is A $\frac{E}{2}$
$E_{i}$ = $\frac{k λ}{r} [sin θ_{1} + sin θ_{2}]$

$E_{i n f i n i t e}$ = $\frac{2 k λ}{r}$
$E_{s e m i - i n f i n i t e}$ = $\frac{k λ}{r}$
$E_{s e m i - i n f i n i t e}$ = $\frac{E}{2}$

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The electric field intensity due to an infinite rod of linear charge density γ at a distance r is E. Then the electric field intensity due to a semi-infinite rod of linear charge density γ at point P, at a perpendicular distance r is

The correct option is A E2 Ei=kλr[sinθ1+sinθ2] Einfinite = 2kλr Esemi−infinite = kλr Esemi−infinite = E2

The electric field intensity due to an infinite rod of linear charge density $γ$ at a distance $r$ is $E$ . Then the electric field intensity due to a semi-infinite rod of linear charge density $γ$ at point $P$ , at a perpendicular distance $r$ is

The correct option is A $\frac{E}{2}$
$E_{i}$ = $\frac{k λ}{r} [sin θ_{1} + sin θ_{2}]$

$E_{i n f i n i t e}$ = $\frac{2 k λ}{r}$
$E_{s e m i - i n f i n i t e}$ = $\frac{k λ}{r}$
$E_{s e m i - i n f i n i t e}$ = $\frac{E}{2}$