Question

Pencil "lead" is actually carbon. What is the current if a $9.0\mathrm{V}$ potential difference is applied between the ends of a $0.70\mathrm{mm}$ diameter, $6.0cm$long lead from a mechanical pencil?

Short answer

If $9.0\mathrm{V}$potential difference is applied between the ends of a $0.70\mathrm{mm}$diameter, $6.0cm$ long lead from a mechanical pencil, then the current is$1.7A$.

Step-by-step solution

Given Information 

Potential difference$=9.0V$

Pencil lead diameter$=0.70mm$

Pencil lead length$=6.0cm$

Explanation

Set $\Delta V=9.0\mathrm{V},d=0.70\mathrm{mm}$and $l=6.0\mathrm{cm}$.

The electric field in the lead is given by

$E=\frac{\Delta V}{l}$

So the current density is,

$J=\sigma E=\sigma \frac{\Delta V}{l}.$

The current is given by

$I=JA,$

Where $A=\frac{1}{4}{r}^2\pi$is the cross-section area of the wire,

$I=\sigma \frac{A}{l}\Delta V$

Modify the expression,

$=\sigma \frac{d^2\pi }{4l}\Delta V$

From Table 27.2 in the book the conductivity of Carbon is $2.9\times {10}^4{\Omega }^{-1}{\mathrm{m}}^{-1}$,

$I=1.7\mathrm{A}$

Final Answer 

Hence, the current is$1.7A$.