The force between two charges when placed in free space is 5 N.If they are in a medium of relative permittivity 5, the force between them willbe

1 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

25 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.5 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 1 N
Fe in free surface
$F e = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}} = 5 N$
In medium
$F e_{m} = \frac{1}{4 π ε_{0} ε_{r}} \frac{q^{2}}{r^{2}}$
$F e_{m} = \frac{F e}{ε_{r}} = \frac{5 N}{5} = 1 N$

Option A is correct.

Suggest Corrections

Similar questions

Two charges each of $100$ micro coulomb are separated in a medium of relative permittivity $2$ by a distance of $5 c m$ . The force between them is :

The magnetic poles placed in air repel each other with a force of $4 \times 10^{- 6} N$ . When the space between them is filled with a medium the repulsive force becomes $2.24 \times 1 0^{- 5} N$ . The relative permeability of the medium is

Q. Two point charges

q_{1}

and

q_{2}

are placed at a distance of

50 m

from each other in air, interact with a certain electrostatic force. If the medium is replaced with oil whose relative permittivity is

5

, then the interaction force between them is still the same. Now the separation between them is

Q. The electric force between two point charges separated by a certain distance in air is

F

the distance at which they should be placed in a medium of relative permittivity

k

so that the force remain the same is:

Q. Two point charges

q_{1}

and

q_{2}

are placed at a distance of 50 m from each other in air, and interact with a certain force. The same charges are now put in oil whose relative permittivity is 5. If the interacting force between them is still the same, their separation now is :

Join BYJU'S Learning Program

The force between two charges when placed in free space is 5 N.If they are in a medium of relative permittivity 5, the force between them willbe

The correct option is A 1 NFe in free surface Fe=14πε0q2r2=5NIn mediumFem=14πε0εrq2r2Fem=Feεr=5N5=1NOption A is correct.

The correct option is A 1 N
Fe in free surface
$F e = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}} = 5 N$
In medium
$F e_{m} = \frac{1}{4 π ε_{0} ε_{r}} \frac{q^{2}}{r^{2}}$
$F e_{m} = \frac{F e}{ε_{r}} = \frac{5 N}{5} = 1 N$

Option A is correct.