Discussion
Two point charges are placed in a medium of relative permittivity 4. The distance at which the force equals what it would be at 1 m in free space is
Answer: A
In a medium the force is divided by the relative permittivity, so at the same separation it would be a quarter of its free space value. To recover the free space force the separation must shrink until the inverse square makes up the factor of four: \[ \frac{1}{\epsilon_r r^{2}} = \frac{1}{\left(1\right)^{2}} \;\Longrightarrow\; r^{2} = \frac{1}{\epsilon_r} = \frac{1}{4} \;\Longrightarrow\; r = 0.5\ \text{m}. \] Whenever a permittivity is traded against a distance the square root appears, because the law is inverse square in one and inverse first power in the other.
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