Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

The surface density of charge on the surface of a charged conductor in the air is $26.5 μ C m^{- 2}$ . Then the outward force per unit area of the charged conductor is $(ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2})$

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Answer:

Solution:

$σ = 26.5 \frac{μ C}{m^{2}} = 26.5 \times 10^{- 6} \frac{C}{m^{2}}$

$\frac{dF}{ds} = \frac{σ^{2}}{2 {kε}_{0}}$

$\frac{dF}{ds} = \frac{26.5 × 26.5 \times 10^{-12}}{2 \times 1 \times 8.85 \times 10^{- 12}}$ $\dots [∵ k = 1 for air]$

$= 39.67 \frac{N}{m^{2}} \approx 39.7 \frac{N}{m^{2}}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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