Should I use 10log or 20log for electric field in decibels?

[tamtam402]

Homework Statement

Let E1 be the electric field 1cm into a person's skin and E0 be the electric field on the surface of that person's skin. If the field loses -486db/m and E0 = 64.1, find the value of E1.

Homework Equations

E1/E0 = k, with k < 1

The Attempt at a Solution

Well I'm mostly confused about the use of 20log vs 10log. My solution is:

We have -486dB/m, or -4.86dB/cm.

10log(E1) = 10log(k) + 10log(E0), where 10log(k) must be negative since k < 1.

We know that 10log(k) = -4.68, thus k = 10^(-4.68/10) = 0.3404

However, my teacher wrote the following equation:
20log(E1) = 20log(k) + 20log(E0)

Which gives a different value of k, obviously. I thought decibels were always 10log(ratio), and the only reason why 20log(ratio) sometimes appear is when you have a ratio of squared values, such as 10log(P1/P2) = 10log([V1^2/R]/[V2^2/r]) = 10log(V1^2/V2^2) = 20log(V1/V2) in electronics.

I'd like to know if 10log or 20log should be used in this situation. Thanks!

 

[tamtam402]

One more thing. Would I be right to assume that using the 20log equations give the ratio k of power density loss that corresponds to the electric field loss? I ask this because the teacher uses his ratio k that he found with the 20log equations, then argues that since P1/P0 = E1^2/E0^2, the ratio of P1/P2 = k^2... I thought that using 20log would already give the "correct k" for the P1/P0 ratio without needing to square it.

 

[tms]

You've basically said it yourself:
$$10\ \log\, \left( \frac{P_1}{P_2} \right) = 10\ \log\, \left( \frac{E_1^2}{E_2^2} \right) = 20\ \log\, \left( \frac{E_1}{E_2} \right). $$
You are using field strength in this problem, so $\ldots$