Question

Obtain ohm’s law from the expression for electrical conductivity.

Answer 1

We can derive the familiar form of Ohm's Law ($V=IR$) by starting with the fundamental expression for electrical conductivity ($\sigma$). This derivation beautifully connects the microscopic behavior of electrons in a material to the macroscopic properties we observe.

The process involves two main stages:
1. Deriving the microscopic (or point) form of Ohm's Law, $J = \sigma E$.
2. Using this to derive the macroscopic form, $V = IR$.

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Starting Point: The Expression for Electrical Conductivity

The electrical conductivity ($\sigma$) of a material is defined by the properties of its charge carriers (usually electrons). The expression, derived from the Drude model, is:

$ \sigma = \frac{ne^2\tau}{m} $

Where:
$n$ = number density of free charge carriers (number of electrons per unit volume, $m^{-3}$)
$e$ = elementary charge (the magnitude of the charge of an electron, $\approx 1.602 \times 10^{-19} C$)
$\tau$ = relaxation time (the average time between collisions for a charge carrier)
$m$ = mass of the charge carrier (mass of an electron)

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Step 1: Deriving the Microscopic Form of Ohm's Law ($J = \sigma E$)

This form relates the current density at a point in a conductor to the electric field at that same point.

1. Force on an Electron: When an external electric field ($E$) is applied across a conductor, each free electron experiences an electrostatic force ($F$):
   $F = eE$

2. Acceleration of an Electron: According to Newton's second law ($F=ma$), this force causes the electron to accelerate:
   $a = \frac{F}{m} = \frac{eE}{m}$

3. Drift Velocity ($v_d$): An electron does not accelerate indefinitely. It constantly collides with the ions of the material's crystal lattice. The drift velocity is the average velocity attained by the electrons due to the electric field, superimposed on their random thermal motion. We can approximate this average velocity as the acceleration multiplied by the average time between collisions ($\tau$):
   $v_d = a \cdot \tau = \left( \frac{eE}{m} \right) \tau = \frac{eE\tau}{m}$

4. Relating Drift Velocity to Current Density ($J$): Current density ($J$) is the amount of current ($I$) flowing through a unit cross-sectional area ($A$). It can be expressed in terms of the charge carriers' properties:
   $J = n e v_d$

5. Combining the Equations: Now, we substitute our expression for drift velocity ($v_d$) into the equation for current density ($J$):
   $J = n e \left( \frac{eE\tau}{m} \right)$

6. Identifying Conductivity: Let's rearrange the terms:
   $J = \left( \frac{ne^2\tau}{m} \right) E$

   Notice that the term in the parentheses is exactly the expression for electrical conductivity, $\sigma$.
   $ \sigma = \frac{ne^2\tau}{m} $

   By substituting $\sigma$ into our equation for $J$, we arrive at the microscopic form of Ohm's Law:

   $ \Large J = \sigma E $

This equation states that the current density in a material is directly proportional to the electric field, and the constant of proportionality is the material's conductivity.

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Step 2: Deriving the Macroscopic Form of Ohm's Law ($V = IR$)

Now, we'll scale up from the microscopic view ($J = \sigma E$) to a macroscopic conductor with length $L$ and uniform cross-sectional area $A$.

1. Relating J and E to I and V:
   * The total current ($I$) flowing through the conductor is the current density ($J$) multiplied by the cross-sectional area ($A$):

   $I = J \cdot A \implies J = \frac{I}{A}$
   

   • For a uniform electric field ($E$) over a length ($L$), the potential difference or voltage ($V$) across the conductor is:
     $V = E \cdot L \implies E = \frac{V}{L}$

2. Substitution: Substitute these macroscopic expressions for $J$ and $E$ back into the microscopic Ohm's Law ($J = \sigma E$):
   $ \frac{I}{A} = \sigma \left( \frac{V}{L} \right) $

3. Rearranging for V: Our goal is to get the familiar $V=IR$ form. Let's solve the equation for $V$:
   $ V = \frac{I \cdot L}{\sigma \cdot A} $

4. Introducing Resistance (R): Let's group the terms related to the material and its geometry:
   $ V = I \left( \frac{L}{\sigma A} \right) $

   We know that resistivity ($\rho$) is the reciprocal of conductivity ($\sigma$):
   $ \rho = \frac{1}{\sigma} $

   Substituting this in, we get:
   $ V = I \left( \rho \frac{L}{A} \right) $

   The term in the parentheses is the definition of resistance ($R$) for a conductor with a uniform cross-section:
   $ R = \rho \frac{L}{A} $

5. Final Result: By substituting $R$ into our equation, we obtain the well-known macroscopic form of Ohm's Law:

   $ \Large V = IR $

Summary

By starting with the fundamental microscopic properties of a material that define its conductivity ($\sigma$), we first derived the relationship between current density ($J$) and the electric field ($E$). Then, by considering a conductor of specific dimensions (length $L$, area $A$), we scaled up this microscopic law to relate the total voltage ($V$) across the conductor to the total current ($I$) flowing through it, successfully obtaining Ohm's Law, $V=IR$.