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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
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
See lessThe 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
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.
See lessWhich 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
Explanation: At the boundary of the dielectric-dielectric, the tangential component of the
electric field intensity is always continuous. We get Et1 = Et2
See lessFind 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
Explanation: At the boundary of a conductor- free space interface, the flux density is
equal to the charge density. Thus D = ρv = 24 units.
See lessFind the electric field if the surface density at the boundary of air is 10 -9 .
c Explanation: It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ = 10 -9 and εo = 10 -9 /36π. We get E = 36π units.
c
Explanation: It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ =
10
-9 and εo = 10
-9
/36π. We get E = 36π units.
See lessFor 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
Explanation: A conservative field implies the work done in a closed path will be zero.
This is given by ∫ E.dl = 0.
See lessThe 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
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.
See lessOn equating the generic form of current density equation and the point form of Ohm’s law, we can obtain V=IR. State True/False
a Explanation: The generic current density equation is J = I/A and the point form of Ohm’s law is J = σ E. On equating both and substituting E = V/L, we get V = IL/σ A = IR which is the Ohm’s law
a
Explanation: The generic current density equation is J = I/A and the point form of Ohm’s
law is J = σ E. On equating both and substituting E = V/L, we get V = IL/σ A = IR which
is the Ohm’s law
See lessCalculate 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
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
See lessFind 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
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.
See less