Two charges q1 and q2 are placed in vacuum at a distance d and force between them is F. If a medium of relative permittivity 4 is introduced between them then new force will be:
Correct answer: A. F/4
- A. F/4
- B. F/2
- C. 2F
- D. 4F
Explanation
The force between two charges in a medium is given by Coulomb's law, which states that the force is inversely proportional to the permittivity of the medium. If the permittivity of the medium increases, the force decreases.In vacuum, F=k⋅q1⋅q2/ d2wherek is Coulomb's constant.When a medium with relative permittivity εr =4 F′=F /εr ( Fmed=Fvac/εr ) =F /4So, the new force will be one-fourth of the original force.
Last updated
About Coulomb's Law
Coulomb's law gives the electrostatic force between two point charges in terms of their magnitudes, separation and the medium between them. Questions involve attraction and repulsion, the inverse square relationship, superposition of forces and vector direction, which must not be confused with the electric field produced by a charge.
Practise Electrostatics
831 free Electrostatics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
According to Coulomb's law, the force between two point charges is proportional to
If a plastic sheet of relative permittivity 2.5 is inserted between two point charges placed in vacuum, then the electrostatic force between them
If the distance between two charges is halved and magnitude of charges are also doubled, then the force between these charges becomes:
If the distance between two identical charges is doubled, then the force of repulsion betweenthem decreases by
If the distance between two point charges is halved, the electrostatic force between them becomes