QJEEM24Phff64: JEE Main 2024 Physics - Class 12 JEEM24Phff64 Solution: Force between two point charges q 1 and q 2 placed in vacuum at …

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QJEEM24Phff64: JEE Main 2024 Physics - Class 12 JEEM24Phff64 Solution: Force between two point charges q 1 and q 2 placed in vacuum at …

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 constant $K = 5$ at ' $r / 5$ ' $cm$ apart will be:

Multiple Choice Options

A 5F
B 25F
C F/5
D F/25

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Accepted Answer

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

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

QJEEM24Phff64: JEE Main 2024 Physics - Class 12 JEEM24Phff64 Solution: Force between two point charges q 1 and q 2 placed in vacuum at …

Multiple Choice Options

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