Test: Electric Field Density
9 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Electric Field Density
| 1 Crore+ students have signed up on EduRev. Have you? |
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: 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.
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: 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.
A uniform surface charge of σ = 2 μC/m2, is situated at z = 2 plane. What is the value of flux density at P(1,1,1)m?
Answer: b
Explanation: The flux density of any field is independent of the position (point). D = σ/2 = 2 X 10-6(-az)/2 = -10-6.
Find the flux density of line charge of radius (cylinder is the Gaussian surface) 2m and charge density is 3.14 units?
Answer: d
Explanation: The electric field of a line charge is given by, E = λ/(2περ), where ρ is the radius of cylinder, which is the Gaussian surface and λ is the charge density. The density D = εE = λ/(2πρ) = 3.14/(2π X 2) = 1/4 = 0.25.
Find the electric field intensity of transformer oil (εr = 2 approx) with density 1/4π (in 109units)
Answer: c
Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units.
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.
If the radius of a sphere is 1/(4π)m and the electric flux density is 16π units, the total flux is given by,
Answer: c
Explanation: Total flux leaving the entire surface is, ψ = 4πr2D from Gauss law. Ψ = 4π(1/16π2) X 16π = 4.
|
11 videos|46 docs|62 tests
|
|
Use Code STAYHOME200 and get INR 200 additional OFF
|
Use Coupon Code |
|
11 videos|46 docs|62 tests
|
Download free EduRev App
Related content with Test: Electric Field Density
|
|
|
|
Related tests with Test: Electric Field Density
|
|
|
|
|
|
© EduRev
|
Education Revolution
|
Follow Us
|
