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.
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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.
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