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
Popular Questions of Class 12th physics
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A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7 C.
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A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
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A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10-12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
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A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
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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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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?
Recently Viewed Questions of Class 12th physics
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Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
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What is the force between two small charged spheres having charges of 2 x 10-7 C and 3 x 10-7 C placed 30 cm apart in air?
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Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf Ω µ?
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(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
Where
is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of 
is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ 
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
[Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
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The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N.
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