# Tag Info

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If you insert a conductor without touching either plates you end up with two capacitors in series with the widths less than that of the original capacitor you had before you inserted the conductor.

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Since many metals form a thin oxide layer of relatively high resistivity, they can be used to affect the dielectric properties of a capacitor made from that material. For example, aluminium, tantalum and even copper all form oxide layers with high resistivity (compared to the respective metals) and high relative permittivity (dielectric constant). Since ...

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It is all about wavelength versus tunnel diameter. The wavelength of GPS is about 20cm it would happily propagate in any normal tunnel if it could get in but the earth and other structures absorb it. AM radio (600kHz - 1500kHz) cannot propagate in any normal tunnel because the wavelength is too long (500m-200m) relative to the diameter, and thus gets ...

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Radio waves are just light of a different frequency. Yes, they are slightly better at reflecting off various surfaces than light, but they still generally travel in straight lines. They don't flow or fill space like a fluid, so there is no reason to believe they'd be good at following you into a tunnel. You generally need line of sight (or something close to ...

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Both of these technologies rely on radio frequency waves, which are blocked by dense matter eg, a hill, a building.

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Force per unit length would be given by $$F=\alpha E.$$ For an infinite line charge the electric field at a distance $d$ is, by Gauss' Law, $$E=\frac{\alpha}{2\pi \epsilon_{0} d}.$$ The dielectric is made of dipoles, so you should be able to figure out why it makes no difference to the field outside the wire. And the two wires are far enough apart that we ...

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You're perfectly correct. Referring to Classical Electrodynamics by Jackson, we see that the index of refraction $n$ is given by: $$n=\sqrt{\frac{\mu}{\mu_{0}}\frac{\epsilon}{\epsilon_{0}}} = \sqrt{\mu_{r}\epsilon_{r}}.$$ But Jackson notes that for most optical frequencies (and non-meta-material media), $\frac{\mu}{\mu_{0}}=\mu_{r}\approx 1$, which is why ...

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