If F is the force between two point charges submerged in a medium of dielectric constant K, then on withdrawing the medium, the force between the charges becomes

Question

If F is the force between two point charges submerged in a medium of dielectric constant K, then on withdrawing the medium, the force between the charges becomes (A) F √K (B) F K (C) (F/√K) (D) (F/K)

Tardigrade Admin · Accepted Answer

F=(1/4 π ε0 K)((q1 q2/r2)), with medium F prime=(1/4 π ε0)((q1 q2/r2)), without medium ⇒ (F prime/F)=K or F prime=F K

Q. If $F$ is the force between two point charges submerged in a medium of dielectric constant $K$ , then on withdrawing the medium, the force between the charges becomes

$F K$

$F K$

$\frac{F}{K}$

$\frac{F}{K}$

$F = \frac{1}{4 π ε _{0} K} (\frac{q _{1} q _{2}}{r ^{2}}),$ with medium
$F^{'} = \frac{1}{4 π ε _{0}} (\frac{q _{1} q _{2}}{r ^{2}}),$ without medium
$\Rightarrow \frac{F ^{'}}{F} = K$ or $F^{'} = F K$

Q. If F is the force between two point charges submerged in a medium of dielectric constant K, then on withdrawing the medium, the force between the charges becomes

FK​

FK

K​F​

KF​

F=4πε0​K1​(r2q1​q2​​), with medium F′=4πε0​1​(r2q1​q2​​), without medium ⇒FF′​=K or F′=FK

Q. If $F$ is the force between two point charges submerged in a medium of dielectric constant $K$ , then on withdrawing the medium, the force between the charges becomes

$F K$

$F K$

$\frac{F}{K}$

$\frac{F}{K}$

$F = \frac{1}{4 π ε _{0} K} (\frac{q _{1} q _{2}}{r ^{2}}),$ with medium
$F^{'} = \frac{1}{4 π ε _{0}} (\frac{q _{1} q _{2}}{r ^{2}}),$ without medium
$\Rightarrow \frac{F ^{'}}{F} = K$ or $F^{'} = F K$