Find the force experienced by a semicircular rod having a charge q as shown in figure. Radius of the wire is R, and the line of charge with linear charge density λ passes through its center and is perpendicular to the plane of wire.

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λq2π2ε0R

λqπ2ε0R

λq4π2ε0R

λq4πε0R

answer is B.

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Detailed Solution

Fnet=∫dq E cos θ=∫−π/2π/2(qπR)Rdθλ2πε0Rcosθ=λq2π2ε0R∫−π/2π/2cosθdθ=λq2π2ε0R[sinθ]−π/2π/2=λq2π2ε0R[1−(−1)]=λqπ2ε0R

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