Relationship between J sigma and E
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Posted by Adnan Raza 7 years, 11 months ago
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Gurvinder Kaur 7 years, 11 months ago
J = sigma E
We know, J = I/A
I = n e A (vd) [in terms of drift velocity of electrons]
Therefore, J = neA(vd)/A = ne (vd)
Now, vd = -e E (tau) /m [ drift velocity of electrons in terms of relaxation time tau]
Therefore, J = n (e square) E (tau)/m
And we know, rho = m / n (e square) (tau)
And 1/rho = sigma = n (e square) (tau) / m
Therefore, J = sigma E
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