Class 12th Physics 2018 Set2 Delhi Board Paper Solution
Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.
Consider a conductor of length L and cross sectional area A , when potental difference v is applied across its end the current produced is i. If n is the number of electrons per unit volume in conductor and Vd is the drift velocity then,
I = -nAeVd Equaltion 1.
Where e is charge on electron , n is no of density
Electric field produced at each point, E = V/L Equation 2
if t is relaxation time and e is the electric field strength then drift velocity
Vd = -etE / m Equation 3
Substituting value in equation 1:
I = - neA(-etE / m )
I = (n e2t / m ) AE Equation 4
As e = V/L then,
I = n e2t A / m x V/L
Or
V/I = m/n e2t x L/A
From other law R = V/I = m/n e2t x L/A Equation 5
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A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?
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A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
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A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
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