1.

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.

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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