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The electric flux density of a surface with permittivity of 2 is given by 12 units. What the flux density of the surface in air?
b Explanation: The relation between electric field and permittivity is given by Dt1/Dt2 = ε1/ε2. Put Dt1 = 12, ε1 = 2 and ε2 =1, we get Dt2 = 12 x 1/ 2 = 6 units.
b
See lessExplanation: The relation between electric field and permittivity is given by Dt1/Dt2 =
ε1/ε2. Put Dt1 = 12, ε1 = 2 and ε2 =1, we get Dt2 = 12 x 1/ 2 = 6 units.
The normal component of which quantity is always discontinuous at the boundary?
b Explanation: The normal component of an electric flux density is always discontinuous at the boundary for a dielectric-dielectric boundary. We get Dn1 = Dn2, when we assume the free surface charge exists at the interface.
b
See lessExplanation: The normal component of an electric flux density is always discontinuous at the boundary for a dielectric-dielectric boundary. We get Dn1 = Dn2, when we assume the free surface charge exists at the interface.
Which component of the electric field intensity is always continuous at the boundary?
a Explanation: At the boundary of the dielectric-dielectric, the tangential component of the electric field intensity is always continuous. We get Et1 = Et2.
a
See lessExplanation: At the boundary of the dielectric-dielectric, the tangential component of the electric field intensity is always continuous. We get Et1 = Et2.
Find the flux density at the boundary when the charge density is given by 24 units.
b Explanation: At the boundary of a conductor- free space interface, the flux density is equal to the charge density. Thus D = ρv = 24 units.
b
See lessExplanation: At the boundary of a conductor- free space interface, the flux density is
equal to the charge density. Thus D = ρv = 24 units.
For a conservative field which of the following equations holds good?
a Explanation: A conservative field implies the work done in a closed path will be zero. This is given by ∫ E.dl = 0.
a
See lessExplanation: A conservative field implies the work done in a closed path will be zero.
This is given by ∫ E.dl = 0.
The charge within a conductor will be
c Explanation: No charges exist in a conductor. An illustration for this statement is that, it is safer to stay inside a car rather than standing under a tree during lightning. Since the car has a metal body, no charges will be possessed by it to get ionized by the lightning.
c
See lessExplanation: No charges exist in a conductor. An illustration for this statement is that, it
is safer to stay inside a car rather than standing under a tree during lightning. Since the
car has a metal body, no charges will be possessed by it to get ionized by the lightning.
Calculate the potential when a conductor of length 2m is having an electric field of 12.3units.
d Explanation: The electric field is given by E = V/L. To get V, put E = 12.3 and L = 2.Thus we get V = E x L = 12.3 x 2 = 24.6 units.
d
See lessExplanation: The electric field is given by E = V/L. To get V, put E = 12.3 and L = 2.Thus we get V = E x L = 12.3 x 2 = 24.6 units.
Find the resistance of a cylinder of area 200 units and length 100m with conductivity of 12 units.
a Explanation: The resistance is given by R = ρL/A = L/σA. Put L = 100, σ = 12 and A = 200, we get R = 100/(12 x 200) = 1/24 units.
a
See lessExplanation: The resistance is given by R = ρL/A = L/σA. Put L = 100, σ = 12 and A =
200, we get R = 100/(12 x 200) = 1/24 units.
Find the mobility of the electrons when the drift velocity is 23 units and the electric field is 11 units.
b Explanation: The mobility is defined as the drift velocity per unit electric field. Thus μe = vd/E = 23/11 = 2.1 units.
b
See lessExplanation: The mobility is defined as the drift velocity per unit electric field. Thus μe = vd/E = 23/11 = 2.1 units.
Compute the conductivity when the current density is 12 units and the electric field is 20 units. Also identify the nature of the material.
c Explanation: The current density is the product of conductivity and electric field intensity. J = σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less than unity, the material is a dielectric.
c
See lessExplanation: The current density is the product of conductivity and electric field intensity. J = σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less than unity, the material is a dielectric.