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PhysicsElectrostaticsConductorMedium2 minQB

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An ellipsoidal cavity is carved within a perfect conductor. A positive charge q is placed at the center of the cavity. The points A & B are on the cavity surface as shown in the figure. Then :

Options:

Answer:

Solution:

$V_{A} = V_{B}$ $r_{A} \geq r_{B}$ so, charge density at $A >$ charge density at B $E_{A} > E_{B}$ flux $= 0$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Conductor

⏱ 2mℹ️ Source: QB

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