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A semi-circular arc of radius a is charged.....
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consider a small portion of the arc now let the small portion be dl so charge dq= λdl but dl = a d@ where @ is teh angle subtended by it at the cnetre thus dq=λ a d@ dE=K dq/a^2 sin@ bcos the cos@ components gets cancelled due to symmetry dE = K λa d@ / a^2 sin@ integrating both sides with limits 0 to E for E and 0 to pie for right hand side E =K λ/a *[cos pie - cos 0] = 2K λ / a |
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