Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceMedium2 minPYQ_2010

PhysicsMediumsingle choice

A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field $E$ at the centre $O$ is

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Solution:

Linear charge density $λ = (\frac{q}{π r})$ $E = \int d E sin θ (- \hat{j}) = \int \frac{K \cdot d q}{r ^{2}} sin θ (- \hat{j})$ $E = \frac{K}{r ^{2}} \int \frac{qr}{π r} d θ sin θ (- \hat{j})$ $= \frac{K}{r ^{2}} \frac{q}{π} \int_{0}^{π} sin θ (- \hat{j})$ $= \frac{q}{2 π ^{2} ε _{0} r ^{2}} (- \hat{j})$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Electric Field & Force

⏱ 2mℹ️ Source: PYQ_2010

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