If the electric field at the centre of the semicircular ring shown in figure. is −xkq^iπR2, then find x.
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Consider an element of length Rdθ and charge on the element dq=λRdθ where λ= charge per unit length. Here sin component of net electric filed due to element of both positive and negative charge will cancel each other and only cosine component contribute the filed. i.e, dEO=2dEcosθ=2kdqR2cosθ=2kλRdθR2cosθ E=2kλRR2∫π20cosθdθ=2kλRR2 Since q=λ(π2)R, thus E=2k(2q)πR2=4qkπR2 As net field in the direction of negative x axis so →E=4qkπR2(−^i) thus, x=4
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