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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceMedium2 minPYQ_2023

PhysicsMediumsingle choice

The electric field intensity on the surface of a solid charged sphere of radius ' $r$ ' and volume charge density ' $ρ$ ' is ( $ε_{0} =$ permittivity of free space $)$

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Answer:

Solution:

According to Gauss' theorem, $ϕ = E \cdot A = \frac{q _{enc}}{ε _{0}}$ But, $q_{enc} = ρ \times$ volume $∴ q_{enc} = ρ (\frac{4}{3} π r^{3})$ And area is $A = 4 π r^{2}$ $∴ E (4 π r^{2}) = \frac{ρ ( \frac{4}{3} π r ^{3} )}{ε _{0}} ∴ E = \frac{ρ r}{3 ε _{0}}$

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

⏱ 2mℹ️ Source: PYQ_2023

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