Force between two charges, when placed in free space is 10 N. If they are in a medium of relative permittivity 5, the force between them will be :

Question

Force between two charges, when placed in free space is 10 N. If they are in a medium of relative permittivity 5, the force between them will be : (A) 2 N (B) 50 N (C) 0.5 N (D) none of the above.

Tardigrade Admin · Accepted Answer

Fm = (1/4 π ε0 εr) (q1 q2/r2) and F0 = (1/4 π ε0) (q1q2/r2) Therefore Fm = (F0/εr) = (10 N/5) = 2 N

Q. Force between two charges, when placed in free space is $10 N$ . If they are in a medium of relative permittivity $5$ , the force between them will be :

2 N

50 N

0.5 N

none of the above.

$F_{m} = \frac{1}{4 π ε _{0} ε _{r}} \frac{q _{1} q _{2}}{r ^{2}}$ and $F_{0} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}}$
Therefore $F_{m} = \frac{F _{0}}{ε _{r}} = \frac{10 N}{5}$ = 2 N

Q. Force between two charges, when placed in free space is 10N. If they are in a medium of relative permittivity 5, the force between them will be :

2 N

50 N

0.5 N

none of the above.

Fm​=4πε0​εr​1​r2q1​q2​​ and F0​=4πε0​1​r2q1​q2​​ Therefore Fm​=εr​F0​​=510N​ = 2 N

Q. Force between two charges, when placed in free space is $10 N$ . If they are in a medium of relative permittivity $5$ , the force between them will be :

$F_{m} = \frac{1}{4 π ε _{0} ε _{r}} \frac{q _{1} q _{2}}{r ^{2}}$ and $F_{0} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}}$
Therefore $F_{m} = \frac{F _{0}}{ε _{r}} = \frac{10 N}{5}$ = 2 N