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Question:

A wire, of length L(=20cm), is bent into a semi-circular arc. If the two equal halves, of the arc, were each to be uniformly charged with charges ±Q, [|Q|=10³ε₀ Coulomb where ε₀ is the permittivity (in SI units) of free space] the net electric field at the centre O of the semi-circular arc would be :

(25×10³N/C)^i

(50×10³N/C)^j

(25×10³N/C)^j

(50×10³N/C)^i

Solution:

Electric field due to quarter ring is(fig)Where K=9×10⁹=1/(4πε₀) and λ=linear charge density=Charge per unit length
By super-position theorem, electric field at centre due to combination of above 2 quarters is given as
→Enet=→E1+→E2
Enet=2Kλ/r^i=2K(2Q/πr)r=4KQ/πr²=4KQ/(π(2L/π)²)
∴Enet=4πKQ/2L²=25×10³N/C^i (Thus option A)