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A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle. |
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Answer» Solution :` dq=Q/LRd THETA ` Net electric FIELD have only vertical component ` =intdEcos theta ` ` =1/(4piepsilon_0)Q/Lint_(-pi/2)^(pi/2)(Rd theta)/(R^2)COS theta ` ` =1/(4piepsilon_0)Q/(LR)[SIN theta]_(-pi/2)^(pi/2) ` ` 1/(4piepsilon_0)Q/(LR)[1-0(-1)] ` ` (2Q)/(4piepsilon_0LR)=Q/(2piepsilonL^2) (:.piR=L)`.
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