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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceMedium2 minQB

PhysicsMediumsingle choice

Figure shows a rod $A B$ , which is bent in a 120° circular arc of radius $R$ . A charge $(- Q)$ is uniformly distributed over rod $A B$ . What is the electric field $E$ at the centre of curvature $O$ ?

Options:

Answer:

Solution:

$λ = \frac{- Q}{Rθ} = \frac{- Q}{R \cdot \frac{2 π}{3}} = \frac{- 3 Q}{2 π R}$

$ε = \frac{2 kλ}{R} sin \frac{θ}{2} = \frac{2 k}{R} \cdot \frac{- 3 Q}{2 π R} sin 6 0^{\circ} (- \hat{i})$

$ε = \frac{3 3 Q}{8 π ^{2} ε _{0} R ^{2}} (+ \hat{i})$ . Option (B).

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

⏱ 2mℹ️ Source: QB

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