A thin semi-circular ring of radius r has a positive charges q distributed uniformly over it. The net field E at the center O is (a) (q)/(4π²ε₀ r²)j (b) -(q)/(4π²ε₀ r²)j (c) -(q)/(2π²ε₀ r²)j (d) (q)/(2π²ε₀ r²)j

Answer: (c). Semicircle: E = (2kλ)/(r) with λ = (q)/(π r), so E = (2q)/(4πε₀π r²) = (q)/(2π²ε₀ r²). The ring lies above O and is positive, so the field at O points down, along - j. So the answer is (c) -(q)/(2π²ε₀ r²) j.

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Electric Field & Field Lines

A thin semi-circular ring of radius $r$ has a positive charges $q$ distributed uniformly over it. The net field $\vec{E}$ at the center $O$ is

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Answer: (c)

Semicircle: $E = \dfrac{2k\lambda}{r}$ with $\lambda = \dfrac{q}{\pi r}$ , so $E = \dfrac{2q}{4\pi\varepsilon_0\pi r^2} = \dfrac{q}{2\pi^2\varepsilon_0 r^2}$ .
The ring lies above O and is positive, so the field at O points down, along $-\hat j$ .
So the answer is (c) $-\dfrac{q}{2\pi^2\varepsilon_0 r^2}\hat j$ .

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