Electrical permittivity

  • Thread starter wangdang
  • Start date
  • #1
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Main Question or Discussion Point

Hi,

I was just wondering if you had an electric field which had the coulombic force acting on the two charges as:
F1= (1/4πε)*(q1q2/d2)
Then what would happen to the force if you placed the charges in a medium of half the electrical permittivity? I tried to solve this using ratios and got:

F1:F2 = (1/4πε)*(q1q2/d2) / (1/4π(1/2)ε)*(q1q2/d2)
= 1/ε / 2 /ε
= 1 / 2

F1/F2 = 1/2
2F1=F2

However this doesn't sound right. If you halved the electrical permittivity, would the force acting on the two charges double?
 

Answers and Replies

  • #2
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[tex] \vec{F} = q \vec{E} = \frac{qq_1\hat{d}}{4\pi \epsilon d^2}[/tex]

If [itex] \epsilon [/itex] is half, the the force is double from my understanding.
 
  • #3
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The force is maximum in empty space. In a dielectric medium the force is reduced due to the electric fields induced in the material.
 
  • #4
35
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permitivity here is not the common sense english term. The word doesnt signify how much does the medium 'permits' the force to act.
In fact higher the permitivity lower the force.
Its true that half permitivity doubles coulumbic force
 

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