Charge Q is distributed uniformly on length l of a wire. It is bent in the form of a semicircular ring. Find the electric field at the centre of the ring.
Q2πε0l2
Q4πε0l2
Qπ4ε0l2
Q2ε0l2
A
Qπ4ε0l2
B
Q4πε0l2
C
Q2ε0l2
D
Q2πε0l2
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Solution
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Consider two small elements of length dl each charge dq on each element. If λ is the line charge density, Q=λl=λ.πr. as wire is bent in the form of semicircular ring. Hence , dq=λdl=Qπr.(rdθ)=Qπdθ. Here sine component of the field at center cancel out and only cosine will contribute the filed. thus, dEc=2dEcosθ=2dq4πϵ0r2cosθ=Q2π2ϵ0r2cosθdθ or Ec=Q2ϵ0(πr)2∫π/20cosθdθ=Q2ϵ0(πr)2(sinπ2−sin0)=Q2ϵ0l2 as (l=πr)
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