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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceMedium2 minQB

PhysicsMediumsingle choice

Two point charges $q_{1}$ and $q_{2}$ are placed at a separation r . The force on $q_{2}$ due to $q_{1}$ is F. Now the complete set-up is placed in a dielectric medium having di-electric constant $k = 4$ . The force due to $q_{1}$ on $q_{2}$ is :

Options:

Answer:

Solution:

In a dielectric medium the Coulomb force reduces by the dielectric constant:

$F^{'} = \frac{F}{k} = \frac{F}{4}$ . Option (D).

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

⏱ 2mℹ️ Source: QB

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