Question

There is a potential difference of 1.1 V between the ends of a 10 cm long graphite rod that has a cross-sectional area of 0.90 mm^2. The resistivity of graphite is 7.5 x 10^-6 Ω-m. (a) Find the resistance of the rod.
(b) Find the current.
(c) Find the electric field inside the rod.

Answer 1

Explanation:

It is given that,

Potential difference between the ends of a rod, V = 1.1 V

Length of the rod, l = 10 cm = 0.1 m

Area of cross section of the rod, $A=0.9\ mm^2=9\times 10^{-7}\ m^2$

The resistivity of graphite, $\rho=7.5\times 10^{-6}\ \Omega-m$

(a) Let R is the resistance of the rod. It is given by :

$R=\rho \dfrac{l}{A}$

$R=7.5\times 10^{-6}\times \dfrac{0.1}{9\times 10^{-7}}$

$R=0.833\ \Omega$

So, the resistance of the rod is 0.833 ohms.

(b) Let I is the current flowing in the wire. It can be calculated using the Ohm's law as :

$I=\dfrac{V}{R}$

$I=\dfrac{1.1\ V}{0.833\ \Omega}$

I = 1.32 A

(c) Let E is the electric field inside the rod. The electric field in terms of potential difference is given by :

$E=\dfrac{V}{l}$

$E=\dfrac{1.1\ V}{0.1\ m}$

E = 11 V/m

Hence, this is the required solution.