Two point charges exert a force F on each other when they are placed r distance apart in air. When they are placed R distance apart in a medium of dielectric constant K, they exert the same force. The distance R equals

Question

Two point charges exert a force F on each other when they are placed r distance apart in air. When they are placed R distance apart in a medium of dielectric constant K, they exert the same force. The distance R equals (A) (r/√K) (B) (r/K) (C) r K (D) r √K

Tardigrade Admin · Accepted Answer

F1=(1/4 π ε0) (q2/r2) or F2=(1/4 π ε0 K) (q2/R2) As F1=F2, hence (q2/r2)=(1/K) (q2/R2) ∴ R=r / √K

Q. Two point charges exert a force $F$ on each other when they are placed $r$ distance apart in air. When they are placed $R$ distance apart in a medium of dielectric constant $K$ , they exert the same force. The distance $R$ equals

$\frac{r}{K}$

$\frac{r}{K}$

$rK$

$r K$

$F_{1} = \frac{1}{4 π ε _{0}} \frac{q ^{2}}{r ^{2}}$ or
$F_{2} = \frac{1}{4 π ε _{0} K} \frac{q ^{2}}{R ^{2}}$
As $F_{1} = F_{2}$ ,
hence $\frac{q ^{2}}{r ^{2}} = \frac{1}{K} \frac{q ^{2}}{R ^{2}}$
$∴ R = r / K$

Q. Two point charges exert a force F on each other when they are placed r distance apart in air. When they are placed R distance apart in a medium of dielectric constant K, they exert the same force. The distance R equals

K​r​

Kr​

rK

rK​

F1​=4πε0​1​r2q2​ or F2​=4πε0​K1​R2q2​ As F1​=F2​, hence r2q2​=K1​R2q2​ ∴R=r/K​

Q. Two point charges exert a force $F$ on each other when they are placed $r$ distance apart in air. When they are placed $R$ distance apart in a medium of dielectric constant $K$ , they exert the same force. The distance $R$ equals

$\frac{r}{K}$

$\frac{r}{K}$

$rK$

$r K$

$F_{1} = \frac{1}{4 π ε _{0}} \frac{q ^{2}}{r ^{2}}$ or
$F_{2} = \frac{1}{4 π ε _{0} K} \frac{q ^{2}}{R ^{2}}$
As $F_{1} = F_{2}$ ,
hence $\frac{q ^{2}}{r ^{2}} = \frac{1}{K} \frac{q ^{2}}{R ^{2}}$
$∴ R = r / K$