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.

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,

The resistivity of graphite,

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

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 = 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 = 11 V/m

Hence, this is the required solution.

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correct it is c.) b b