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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

Two similar point charges $q_{1}$ and $q_{2}$ are placed at a distance $r$ apart in the air. The force between them is $F_{1}$ . A dielectric slab of thickness $t (< r)$ and dielectric constant $K$ is placed between the charges. Then the force between the same charges is $F_{2}$ . The ratio is

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Answer:

Solution:

$F_{1} = \frac{k q_{1} q_{2}}{r^{2}} = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}, F_{2} = \frac{q_{1} q_{2}}{4 π ε_{0} (r - t + t \sqrt{K})^{2}}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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