Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceEasy2 minPYQ_2019

PhysicsEasysingle choice

A parallel plate capacitor with plates of area $1 m^{2}$ each, are at a separation of $0.1 m .$ If the electric field between the plates is $100 N / C,$ the magnitude of charge on each plate is: $(Take \in_{0} = 8.85 \times 10^{- 12} \frac{C^{2}}{N - m^{2}})$

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Answer:

Solution:

$E = \frac{σ}{ε_{0}} = \frac{Q}{A ε_{0}} = 100 V / m$

$Q = 100 \times A ε_{0}$

$\Rightarrow Q = 100 \times 1 \times 8.85 \times 10^{- 12} C$

$= 8.85 \times 10^{- 10} C$

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

⏱ 2mℹ️ Source: PYQ_2019

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