# Homework Help: Units help

1. Nov 6, 2008

### tony873004

Help with units
The book gives the following forumla
$$E_0 = \sqrt {2\mu _0 c\left\langle S \right\rangle }$$

And when it uses it in an example, the units go from:
$$\sqrt {\left( {{\rm{H/m}}} \right)\left( {{\rm{m/s}}} \right)\left( {{\rm{W/m}}^{\rm{2}} } \right)}$$
to $$V/m$$, with no intermediate steps. My two efforts fall short:

Attempt #1
$$\sqrt {\left( {{\rm{H/m}}} \right)\left( {{\rm{m/s}}} \right)\left( {{\rm{W/m}}^{\rm{2}} } \right)} = \sqrt {\frac{{{\rm{HmW}}}}{{{\rm{msmm}}}}} = \sqrt {\frac{{{\rm{HW}}}}{{{\rm{m}}^{\rm{2}} {\rm{s}}}}}$$
Does a Henry - Watt divided by a second equal a Volt squared? If so, = V/m

Attempt #2 using N/A^2 instead of H/m for units of mu0
$$\begin{array}{l} \sqrt {\left( {{\rm{N/A}}^{\rm{2}} } \right)\left( {{\rm{/s}}} \right)\left( {{\rm{W/}}{\rm{m}}} \right)} = \\ \\ \sqrt {\frac{{{\rm{NW}}}}{{{\rm{A}}^{\rm{2}} {\rm{s}} \cdot {\rm{m}}}}} = \sqrt {\frac{{{\rm{kgmJ}}}}{{{\rm{A}}^{\rm{2}} {\rm{s}} \cdot {\rm{ms}}^{\rm{2}} {\rm{s}}}}} = \sqrt {\frac{{{\rm{kg - m - kg - m - m}}}}{{\left( {\frac{{\rm{C}}}{{\rm{s}}}} \right)^{\rm{2}} {\rm{s - m - s}}^{\rm{2}} {\rm{s - s}}^{\rm{2}} }}} = \sqrt {\frac{{{\rm{kg - m - kg - m - }}{\rm{ - }}}}{{{\rm{C}}^{\rm{2}} {\rm{s - }}{\rm{ - s}}^{\rm{2}} {\rm{ - s - }}}}} \\ \\ = \sqrt {\frac{{{\rm{kg - m - kg - m}}}}{{{\rm{C}}^{\rm{2}} {\rm{s - s}}^{\rm{2}} {\rm{ - s}}}}} = \sqrt {\frac{{{\rm{kg - m - kg - m}}}}{{\left( {{\rm{FV}}} \right)^{\rm{2}} {\rm{s - s}}^{\rm{2}} {\rm{ - s}}}}} = \sqrt {\frac{{{\rm{kg - m - kg - m}}}}{{{\rm{F}}^{\rm{2}} {\rm{V}}^{\rm{2}} {\rm{ - s - s}}^{\rm{2}} {\rm{ - s}}}}} = \sqrt {\frac{{{\rm{NW}}}}{{{\rm{A}}^{\rm{2}} {\rm{s - m}}}}} = \sqrt {\frac{{{\rm{kg - }}{\rm{ - mJ}}}}{{{\rm{A}}^{\rm{2}} {\rm{s}}{\rm{ - s - s}}^{\rm{2}} }}} = \\ \\ \sqrt {\frac{{{\rm{kg - mJ}}}}{{{\rm{A}}^{\rm{2}} {\rm{s - s - s}}^{\rm{2}} }}} \\ \end{array}$$

This is not beginning to resemble $$\sqrt {\frac{{{\rm{V}}^{\rm{2}} }}{{{\rm{m}}^2 }}}$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 6, 2008

### D H

Staff Emeritus
Henries have the same units as do volt-seconds/ampere. Watts are of course volt-amperes. Thus (H/m)·(m/s)·(W/m2) is equivalent to (V·s)/(A·m)·(m/s)·(V·A/m2) = (V/A)·(V·A/m2) = (V/m)2

3. Nov 6, 2008

### joeyar

An equation that may come in handy:

L*dI/dt = V.

Since we're doing a dimensional analysis, we can ignore the (dimensionless) d's"

LI/t = V.

Multiplying both sides by V, LIV/t = V^2.

Power = P = IV, so substituting this in we get LP/t = V^2.

The unit of L is H (inductance is measured in Henrys), the unit of P is W (power is measured in watts) and the unit of t is s, so it is indeed true that a volt squared = henry times a watt (or a henry-watt) divided by a second.

4. Nov 6, 2008

### tony873004

Thanks, DH and Joeyar. I doubt I was required to justify this step since the book doesn't, but I just had to know :)