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Electrostatics Physics Electrostatic Potential and Capacitance Physics Class 12th

The force between two charged particle placed in air at separation x is F₀. Both the charged particle immerged in a medium of dielectric constant K without changing separation between two charge, then net force on one of the particle is now

Option 1 -

$\frac{F_{0}}{K}$

Option 2 -

$\frac{F_{0}}{2 K}$

Option 3 -

$\frac{2 F_{0}}{K}$

Option 4 -

$F_{0}$

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

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

In medium $F' = \frac{1}{4 π (k \in_{0})} \frac{q_{1} q_{2}}{r^{2}}$

$F' = \frac{F_{0}}{K}$

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The force between two charged particle placed in air at separation x is F₀. Both the charged particle immerged in a medium of dielectric constant K without changing separation between two charge, then net force on one of the particle is now

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The force between two charged particle placed in air at separation x is F0. Both the charged particle immerged in a medium of dielectric constant K without changing separation between two charge, then net force on one of the particle is now

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The force between two charged particle placed in air at separation x is F₀. Both the charged particle immerged in a medium of dielectric constant K without changing separation between two charge, then net force on one of the particle is now