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Find the flux density of a sheet of charge density 25 units in air.
Answer: b Explanation: Electric field intensity of infinite sheet of charge E = σ/2ε. Thus D = εE = σ/2 = 25/2 = 12.5.
Answer: b
See lessExplanation: Electric field intensity of infinite sheet of charge E = σ/2ε.
Thus D = εE = σ/2 = 25/2 = 12.5.
The Gaussian surface is
Answer: b Explanation: It is any physical or imaginary closed surface around a charge which satisfies the following condition: D is everywhere either normal or tangential to the surface so that D.ds becomes either Dds or 0 respectively.
Answer: b
See lessExplanation: It is any physical or imaginary closed surface around a charge which
satisfies the following condition: D is everywhere either normal or tangential to the
surface so that D.ds becomes either Dds or 0 respectively.
Which of the following correctly states Gauss law?
Answer: d Explanation: The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. In other words, electric flux per unit volume leaving a point (vanishing small volume), is equal to the volume charge density.
Answer: d
See lessExplanation: The electric flux passing through any closed surface is equal to the total
charge enclosed by that surface. In other words, electric flux per unit volume leaving a
point (vanishing small volume), is equal to the volume charge density.
The electric flux density is the
Answer: a Explanation: D= εE, where ε=εoεr is the permittivity of electric field and E is the electric field intensity. Thus electric flux density is the product of permittivity and electric field intensity.
Answer: a
See lessExplanation: D= εE, where ε=εoεr is the permittivity of electric field and E is the electric field intensity. Thus electric flux density is the product of permittivity and electric field intensity.
Electric flux density in electric field is referred to as
Answer: b Explanation: Electric flux density is given by the ratio between number of flux lines crossing a surface normal to the lines and the surface area. The direction of D at a point is the direction of the flux lines at that point.
Answer: b
See lessExplanation: Electric flux density is given by the ratio between number of flux lines
crossing a surface normal to the lines and the surface area. The direction of D at a point is the direction of the flux lines at that point.
The lines of force are said to be
Answer: c Explanation: The lines drawn to trace the direction in which a positive test charge will experience force due to the main charge are called lines of force. They are not real but drawn for our interpretation.
Answer: c
See lessExplanation: The lines drawn to trace the direction in which a positive test charge will
experience force due to the main charge are called lines of force. They are not real but
drawn for our interpretation.
In electromagnetic waves, the electric field will be perpendicular to which of the following?
Answer: c Explanation: In an electromagnetic wave, the electric field and magnetic field will be perpendicular to each other. Both of these fields will be perpendicular to the wave propagation.
Answer: c
See lessExplanation: In an electromagnetic wave, the electric field and magnetic field will be
perpendicular to each other. Both of these fields will be perpendicular to the wave
propagation.
For a test charge placed at infinity, the electric field will be
Answer: c Explanation: E = Q/ (4∏εor2) When distance d is infinity, the electric field will be zero, E= 0.
Answer: c
See lessExplanation: E = Q/ (4∏εor2)
When distance d is infinity, the electric field will be zero, E= 0.
Electric field intensity due to infinite sheet of charge σ is
Answer: d Explanation: E = σ/2ε.(1- cos α), where α = h/(√(h2+a2)) Here, h is the distance of the sheet from point P and a is the radius of the sheet. For infinite sheet, α = 90. Thus E = σ/2ε.
Answer: d
See lessExplanation: E = σ/2ε.(1- cos α), where α = h/(√(h2+a2))
Here, h is the distance of the sheet from point P and a is the radius of the sheet. For
infinite sheet, α = 90. Thus E = σ/2ε.
Electric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.
Answer: a Explanation: The electric field intensity of an infinitely long conductor is given by, E = λ/(4πεh).(sin α2 – sin α1)i + (cos α2 + cos α1)j For an infinitely long conductor, α = 0. E = λ/(4πεh).(cos 0 + cos 0) = λ/(2πεh).aN
Answer: a
See lessExplanation: The electric field intensity of an infinitely long conductor is given by, E =
λ/(4πεh).(sin α2 – sin α1)i + (cos α2 + cos α1)j
For an infinitely long conductor, α = 0. E = λ/(4πεh).(cos 0 + cos 0) = λ/(2πεh).aN