Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsGauss LawEasy2 minPYQ_2020
PhysicsEasysingle choice

The sketch below shows cross-sections of the equipotential surface between two charged conductors that are shown in solid black. Some points on the equipotential surfaces near the conductors are marked asA,B,C.......... The arrangement lies in the air.Take ϵ0=8.85×10-12 C2 N-1 m-2
The surface charge density of the plate is equal to

Options:

Answer:
A
Solution:

E=40-100.3=100 V m-1

(near the plate the electric field has to be uniform

∴  it is almost due to the plate).

For conducting plate

E=σε0 ⇒    σ=ε0E

Therefore     σ=8.85×10-12×100

= 8.85×10-10 C m-2

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Gauss Law
⏱ 2mℹ️ Source: PYQ_2020

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