the force between two point charge separated by air is 4N. When separated by a medium of relative permittivity 2,the force between them becomes
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the force between two point charge separated by air is 4N. When separated by a medium of relative permittivity 2,the force between them becomes
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Answer:
2N
Explanation:
F = 1/(4π k∈₀) q₁q₂/r²
When in air, k∈₀ has its original value for air.
F = 1/(4π∈) q₁q₂/r²
when, k = 2.
F' = 1/(4π 2∈₀) q₁q₂/r²
= 1/2 * 1/(4π∈) q₁q₂/r²
= 1/2 * F
= 1/2 * 4N = 2N
Answer:
Force between the 2 charges becomes 2N.
Explanation:
Given that the force between 2 point charges separated by the air is 4N.
The permittivity of the air is considered as 1.
So accordingto coulomb's law of force,
[tex]F=\frac{Q_{1} Q_{2} }{4\pi E_{0}r^{2} }[/tex]
When the same system is transferred to a medium of relative permittivity
of 2 ([tex]E_{r}=2[/tex]) .
The Coulombic Force realation becomes
[tex]F_{medium}=\frac{Q_{1} Q_{2} }{4\pi E_{0}E_{r} r^{2} } \\[/tex]
[tex]F_{medium}=\frac{Q_{1} Q_{2} }{4\pi E_{0}2r^{2} }[/tex]
[tex]F_{medium}=\frac{F}{2}=\frac{4}{2} =2N[/tex]
Hence, The force between the two charges changes from 4N to 2N.