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Chapter 27: Q. 33 (page 763)

Pencil "lead" is actually carbon. What is the current if a 9.0V potential difference is applied between the ends of a 0.70mm diameter, 6.0cmlong lead from a mechanical pencil?

Short Answer

Expert verified

If 9.0Vpotential difference is applied between the ends of a 0.70mmdiameter, 6.0cm long lead from a mechanical pencil, then the current is1.7A.

Step by step solution

01

Given Information 

Potential difference=9.0V

Pencil lead diameter=0.70mm

Pencil lead length=6.0cm

02

Explanation

Set ΔV=9.0V,d=0.70mmand l=6.0cm.

The electric field in the lead is given by

E=ΔVl

So the current density is,

J=σE=σΔVl.

The current is given by

I=JA,

Where A=14r2πis the cross-section area of the wire,

I=σAlΔV

Modify the expression,

=σd2π4lΔV

From Table 27.2 in the book the conductivity of Carbon is 2.9×104Ω-1m-1,

I=1.7A

03

Final Answer 

Hence, the current is1.7A.

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