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Standard XII

Physics

Coulomb's Law - Mathematical Form

Two charges a...

Question

Two charges are placed in vacuum at a distance 'd' apart. The force between them is 'F'. If a medium of a dielectric constant 4 is introduced between them, the force will be:

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F/2

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F/4

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Solution

The correct option is D F/4
$F = \frac{2}{4 π ϵ_{0} ϵ_{r}} \frac{q_{1} q_{2}}{F d^{2}}$
$F α \frac{1}{ϵ_{r}}$
For vacuum $ϵ_{r} = 1,$
$ϵ_{r}$ of medium $= 4$
$∴$ F for medium will be F/4.

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Two charges are placed in vacuum at a distance 'd' apart. The force between them is 'F'. If a medium of a dielectric constant 4 is introduced between them, the force will be:

The correct option is D F/4F=24πϵ0ϵrq1q2Fd2Fα1ϵrFor vacuum ϵr=1,ϵr of medium =4∴ F for medium will be F/4.

The correct option is D F/4
$F = \frac{2}{4 π ϵ_{0} ϵ_{r}} \frac{q_{1} q_{2}}{F d^{2}}$
$F α \frac{1}{ϵ_{r}}$
For vacuum $ϵ_{r} = 1,$
$ϵ_{r}$ of medium $= 4$
$∴$ F for medium will be F/4.