The conducting path between the right hand and the left hand can be modeled as a 10-cm-diameter, 160-cm-long cylinder. The average resistivity of the interior of the human body is 5.0 Ω m. Dry skin has a much higher resistivity, but skin resistance can be made negligible by soaking the hands in salt water. If skin resistance is neglected, what potential difference between the hands is needed for a lethal shock of 100 mA across the chest? Your result shows that even small potential differences can produce dangerous currents when the skin is wet.

Given in the question,

Set $d=10\mathrm{cm},l=160\mathrm{cm}$and $\rho =5.0\Omega \mathrm{m}$, and $I=100\mathrm{mA}$.

The area of the cross-section of the cylinder that is used to model the route between your hands is

$A=\frac{1}{4}\pi d^2$

The resistance of this cylinder is

$R=\rho \frac{l}{A}=\rho \frac{4l}{\pi d^2}$

The current and the potential difference are related by Ohm's law

$I=\frac{\Delta V}{R}$

Calculate the resistivity,

$R=\frac{\rho L}{A}$

The expression for the potential difference,

$\mathrm{\Delta }V=\frac{4\rho IL}{\pi d^2}$

The diameter of the arms is $10\mathrm{cm}$.

Convert units of diameter from $cm$to $m,$

$d=10\mathrm{cm}$

$=(10\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=0.1\mathrm{m}$

The length of arm is $160\mathrm{cm}$.

Convert units of length from $cm$to $m,$

$L=160\mathrm{cm}$

$=(160\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=1.6\mathrm{m}$

Current across the chest is $100\mathrm{mA}$.

Convert units of current from $\mathrm{mA}$to $\mathrm{A}$.

$I=100\mathrm{mA}$

$=(100\mathrm{mA})\left(\frac{1\mathrm{A}}{1000\mathrm{mA}}\right)$

$=0.1\mathrm{A}$

Substitute $\frac{4\rho L}{\pi d^2}$for $R$in the equation $\Delta V=IR$and solve for $∆V,$

$\Delta V=I\left(\frac{4\rho L}{\pi d^2}\right)$

$=\frac{4\rho IL}{\pi d^2}$

Substitute the values,

role="math" localid="1649150118248" $\mathrm{\Delta }V=\left(\frac{4(5.0\mathrm{\Omega }\cdot \mathrm{m})\left(0.100\mathrm{A}\right)\left(1.60\mathrm{m}\right)}{\pi (0.10\mathrm{m}{)}^2}\right)$

$=101.86\mathrm{V}$

Therefore, the potential difference required for lethal shock is$101.86V$