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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceMedium2 minQB

PhysicsMediummultiple choice

A non conducting ring of radius R is charged as shown in figure :

Options:(select one or more)

Answer:

B, D

Solution:

The two halves carry +q and -q, so at the centre $V = \frac{q - q}{4 π ε _{0} R} = 0$ (B correct, C wrong).

Each semicircle gives $E_{1} = \frac{λ}{2 π ε _{0} R} = \frac{q}{2 π ^{2} ε _{0} R ^{2}}$ ; both fields point the same way: $E = \frac{q}{π ^{2} ε _{0} R ^{2}}$ (D correct, A wrong).

Subject:PhysicsTopic:ElectrostaticsSubtopic:Electric Field & Force

⏱ 2mℹ️ Source: QB

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