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Electric field at centre O of semicircle of radius 'a' having linear charge density $\lambda$ is given by
A
$\dfrac{2\lambda}{\varepsilon_0 a}$
B
$\dfrac{\lambda\pi}{\varepsilon_0 a}$
C
$\dfrac{\lambda}{2\pi\varepsilon_0 a}$
D
$\dfrac{\lambda}{\pi\varepsilon_0 a}$
Detailed Solution
For a charged circular arc subtending angle $\alpha$ at the centre, the field at the centre is $E = \dfrac{2K\lambda}{a}\sin\dfrac{\alpha}{2}$ (the components along the arc's symmetry axis add, and the perpendicular components cancel).
[add image: semicircular arc of radius a with centre O, field at O along the symmetry axis]
For a semicircle, $\alpha = \pi$, so $\sin\dfrac{\pi}{2} = 1$:
$E = \dfrac{2K\lambda}{a} = \dfrac{2\lambda}{4\pi\varepsilon_0 a}$
$E = \dfrac{\lambda}{2\pi\varepsilon_0 a}$
[add image: semicircular arc of radius a with centre O, field at O along the symmetry axis]
For a semicircle, $\alpha = \pi$, so $\sin\dfrac{\pi}{2} = 1$:
$E = \dfrac{2K\lambda}{a} = \dfrac{2\lambda}{4\pi\varepsilon_0 a}$
$E = \dfrac{\lambda}{2\pi\varepsilon_0 a}$