Two point charges placed at a distance r in air experience a certain force. Then the distance at which they will experience the same force in a medium of dielectric constant K is

Question

Two point charges placed at a distance r in air experience a certain force. Then the distance at which they will experience the same force in a medium of dielectric constant K is (A) r/K (B) Kr (C) (r/√K) (D) r+√K

Tardigrade Admin · Accepted Answer

(q1 q2/r2)=(q1 q2/K r' 2) or K r' 2=r2 or r'=(r/√K)

Q. Two point charges placed at a distance $r$ in air experience a certain force. Then the distance at which they will experience the same force in a medium of dielectric constant $K$ is

r/K

Kr

$\frac{r}{K}$

$r + K$

$\frac{q _{1} q _{2}}{r ^{2}} = \frac{q _{1} q _{2}}{K r ^{^{'} 2}}$ or $K r^{^{'} 2} = r^{2}$ or $r^{'} = \frac{r}{K}$

Q. Two point charges placed at a distance r in air experience a certain force. Then the distance at which they will experience the same force in a medium of dielectric constant K is

r/K

Kr

K​r​

r+K​

r2q1​q2​​=Kr′2q1​q2​​ or Kr′2=r2 or r′=K​r​

Q. Two point charges placed at a distance $r$ in air experience a certain force. Then the distance at which they will experience the same force in a medium of dielectric constant $K$ is

$\frac{r}{K}$

$r + K$

$\frac{q _{1} q _{2}}{r ^{2}} = \frac{q _{1} q _{2}}{K r ^{^{'} 2}}$ or $K r^{^{'} 2} = r^{2}$ or $r^{'} = \frac{r}{K}$