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

Step-by-step 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* 10^(-7)\ m^2`

The resistivity of graphite,
`\rho=7.5* 10^(-6)\ \Omega-m`

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

`R=\rho (l)/(A)`

`R=7.5* 10^(-6)* (0.1)/(9* 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=(V)/(R)`

`I=(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=(V)/(l)`

`E=(1.1\ V)/(0.1\ m)`

E = 11 V/m

Hence, this is the required solution.