Question 782: Electrostatics & Capacitance

Question

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

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
$$E = \int_{ - \pi /2}^{\pi /2} {dE} $$
$$ = 2\int_O^{\pi /2} {{1 \over {4\pi {\varepsilon _0}}}{{\lambda \left( {ad	heta } ight)} \over {{a^2}}}\cos 	heta } $$
$$ = 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda  \over a}\left[ {\sin 	heta } ight]_o^{\pi /2}$$
$$ = 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda  \over a}.1 = {\lambda  \over {2\pi {\varepsilon _0}a}}$$

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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