Question
What is specific conductivity ? Give its unit in S.I. system. Prove that $ \vec{j}=\sigma\vec{E} $, where $ \vec{E} = $ Intensity of electric field, $ \vec{j} = $ current density and $ \sigma = $ specific conductivity.
✓
Answer
Specific Conductivity : The reciprocal of specific resistance of a material is called specific conductivity. It is denoted by $ \sigma $.
∴ Specific conductivity $=\frac{1}{\text { Specific Resistance }}$ i.e. $\sigma=\frac{1}{\rho}$
Proof of $ \vec{j} = \sigma \cdot \vec{E} $ : When an electric potential difference V is applied across the ends of an electric conductor of length l and area of cross-section A, then an electric field $ E = V/l $ is established inside the conductor. In this field, free electrons in the conductor begin to drift with drift velocity $ v_d $ due to which electric current I flowing through the conductor is given by: $ I = ne A v_d $ and
$ v_d = \frac{eV\tau}{ml} \Rightarrow v_d = e(\frac{V}{l})\frac{\tau}{m} \Rightarrow \frac{eE\tau}{m} $
$\therefore I = neA(\frac{eE\tau}{m}) = \frac{ne^2AE\tau}{m} $
$ \frac{I}{A} = (\frac{ne^2\tau}{m})E $
But $ \frac{I}{A} = current~density = j $
and $ \frac{m}{ne^2\tau} = \rho \Rightarrow \frac{ne^2\tau}{m} = \frac{1}{\rho} $
$ \Rightarrow j = (\frac{1}{\rho})E $ but $ \frac{1}{\rho} = \sigma $
$ j = \sigma E $ or in vector form $ \vec{j} = \sigma \cdot \vec{E} $
∴ Specific conductivity $=\frac{1}{\text { Specific Resistance }}$ i.e. $\sigma=\frac{1}{\rho}$
Proof of $ \vec{j} = \sigma \cdot \vec{E} $ : When an electric potential difference V is applied across the ends of an electric conductor of length l and area of cross-section A, then an electric field $ E = V/l $ is established inside the conductor. In this field, free electrons in the conductor begin to drift with drift velocity $ v_d $ due to which electric current I flowing through the conductor is given by: $ I = ne A v_d $ and
$ v_d = \frac{eV\tau}{ml} \Rightarrow v_d = e(\frac{V}{l})\frac{\tau}{m} \Rightarrow \frac{eE\tau}{m} $
$\therefore I = neA(\frac{eE\tau}{m}) = \frac{ne^2AE\tau}{m} $
$ \frac{I}{A} = (\frac{ne^2\tau}{m})E $
But $ \frac{I}{A} = current~density = j $
and $ \frac{m}{ne^2\tau} = \rho \Rightarrow \frac{ne^2\tau}{m} = \frac{1}{\rho} $
$ \Rightarrow j = (\frac{1}{\rho})E $ but $ \frac{1}{\rho} = \sigma $
$ j = \sigma E $ or in vector form $ \vec{j} = \sigma \cdot \vec{E} $
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