Question 782: Electrostatics & Capacitance

Question

Electric field at centre O of semicircle of radius $a$ having linear charge density $\lambda $ given as
![](https://neetcare-image.cdn.examgoal.net/1qpqdn5k7tjn2kh/8f2b8578-af31-40ef-a2dd-a47665b99bee/e1da0a20-6702-11ea-997e-790b4402a570/file-1qpqdn5k7tjn2ki-165w.jpg)

Accepted Answer

Considering symmetric elements each of length dl at A and B, we note that electric fields perpendicular to PO are cancelled and those along PO are added. The electric field due to an element of length $dl( = ad	heta)$ along PO.
$dE = {1 \over {4\pi {\varepsilon _0}}}{{dq} \over {{a^2}}}\cos 	heta $
$\left(\because {dl = ad	heta } ight)$
$ = {1 \over {4\pi {\varepsilon _0}}}{{\lambda dl} \over {{a^2}}}\cos 	heta $
$ = {1 \over {4\pi {\varepsilon _0}}}{{\lambda \left( {ad	heta } ight)} \over {{a^2}}}\cos 	heta $ Net electric field at O
$E = \int_{ - \pi /2}^{\pi /2} {dE} $
$ = 2\int_O^{\pi /2} {{1 \over {4\pi {\varepsilon _0}}}{{\lambda \left( {ad	heta } ight)} \over {{a^2}}}\cos 	heta } $
$ = 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda  \over a}\left[ {\sin 	heta } ight]_o^{\pi /2}$
$ = 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda  \over a}.1 = {\lambda  \over {2\pi {\varepsilon _0}a}}$
![](https://neetcare-image.cdn.examgoal.net/1qpqdn5k7tk1g5p/29db2ca3-bb3a-429c-9396-08b742b62e41/71ad62e0-6704-11ea-997e-790b4402a570/file-1qpqdn5k7tk1g5q-222w.jfif)

Answer

${{2\lambda } \over {{\varepsilon _0}a}}$

Answer

${{\lambda \pi } \over {{\varepsilon _0}a}}$

Answer

${\lambda  \over {2\pi {\varepsilon _0}a}}$

Answer

${\lambda  \over {\pi {\varepsilon _0}a}}$

Electrostatics & Capacitance

Electric field at centre O of semicircle of radius a having linear charge density \lambda given as

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