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PhysicsElectrostaticsElectric Field & ForceEasy2 minPYQ_2024

PhysicsEasysingle choice

Force between two point charges $q_{1}$ and $q_{2}$ placed in vacuum at $r$ $cm$ apart is $F$ . Force between them when placed in a medium having dielectric $K = 5$ at $\frac{r}{5}$ $cm$ apart will be:

Options:

Answer:

Solution:

Using Coulomb's law, we can write:

In air $F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$

In medium $F' = \frac{1}{4 π (K ε_{0})} \frac{q_{1} q_{2}}{{(r')}^{2}} = \frac{1}{4 π (5 ε_{0})} \frac{25 q_{1} q_{2}}{{(r)}^{2}} = 5 F$

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

⏱ 2mℹ️ Source: PYQ_2024

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