First slide
Electric field
Question

A semicircular wire is uniformly charged with linear charge density dependent on the angle θ from y-direction as λ=λ0|sinθ|, where λ0 is a constant. The electric field intensity at the centre of the arc is

Difficult
Solution

Lets take an element at angle dθ as shown. Electric field due to this element is

dE=dqR2

Horizontal component dE sin θ will be cancelled out with opposite element, and dE cosθ components get added up.

 So E=θ=π/2θ=π/2dEcosθ=2θ=0θ=π/2Kλ0sinθR2RdθcosθE=λ04πε0R(j^)

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