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We know that electricity cannot pass through glass and wood. but what if we decrease their resistance? We can do so by taking a very broad cylinder (high diameter) because $$R= \frac{\rho L}{A}$$

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  • $\begingroup$ It's $\rho$ that defines the conductivity. $\endgroup$ – user36790 Nov 12 '15 at 7:22
  • $\begingroup$ I think you meant short pipe? So that $L$ is low? $\endgroup$ – SchrodingersCat Nov 12 '15 at 7:26
  • $\begingroup$ Just do the math using the formula you posted. This is the difference in magnitude between the resistivity of glass and that of copper: $\rho_{glass} = 10^{11} \Omega m$, $\rho_{Cu} = 1.68 * 10^{-8} \Omega m$. $\endgroup$ – Energizer777 Nov 12 '15 at 7:29
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    $\begingroup$ Its worth noting that most equations, such as the one given, are only approximations to reality, and if you were to create an extreme situation such as a really thin pipe (like an atom wide), or a really short pipe, the equation breaks down. $\endgroup$ – Joshua Lin Nov 12 '15 at 7:50
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    $\begingroup$ It isn't clear what you're asking. The equation you give is correct and subject to the restrictions mentioned by Joshua will give the correct result for the resistance of a piece of glass or wood. However it isn't clear what precisely you're asking. $\endgroup$ – John Rennie Nov 12 '15 at 8:05
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Lets do the arithmetic, as suggested by Energizer777

$$R= \frac{\rho L}{A}$$ $$\rho_{copper} = 10^{-8} \Omega m$$ $$\rho_{glass} = 10^{11} \Omega m$$

How wide a piece of glass would I need to have resistance (per meter length) equal to a very fine copper wire with a radius of 0.1 mm?

The area of my copper wire is $\pi r^2 = 3.14 \times 10^{-8} m^2$

The ratio of $\rho_{glass} : \rho_{copper}$ is $10^{19}$

Consequently the area of glass I need is $3.14 \times 10^{11} m^2$

This would have a radius of $316 km$.

In practice it would be impossible to deliver current uniformly to all parts of the front face of such a large object. I note that resting on the ground, the thickness of your glass "wire" would extend far into into space. Constructing such an object would be impossible on Earth (due to gravity).

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We know that electricity cannot pass through glass and wood

There is no such thing as a perfect insulator. There is always some minute current flowing through any insulator, including glass and wood.

There are ways in which the resistance of an insulator can be reduced. One way is the one you mentioned: a very wide block of it will conduct more than a narrow wire. Resistance can also be decreased (conductivity increased) by other means. For example, wood will be more conductive when it is moist than when it is completely dry, as the water will allow ions in the wood to move, and hence carry current. Different woods also have different conductivities.

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