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The Gaussian surface for a line charge will be
Answer: b Explanation: A line charge can be visualized as a rod of electric charges. The three dimensional imaginary enclosed surface of a rod can be a cylinder.
Answer: b
Explanation: A line charge can be visualized as a rod of electric charges. The three
dimensional imaginary enclosed surface of a rod can be a cylinder.
See lessThe Gaussian surface for a point charge will be
Answer: c Explanation: A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.
Answer: c
Explanation: A point charge is single dimensional. The three dimensional imaginary
enclosed surface of a point charge will be sphere.
See lessA circular disc of radius 5m with a surface charge density ρs = 10sinφ is enclosed by surface. What is the net flux crossing the surface?
Answer: d Explanation: Q = ∫ ρsds = ∫∫ 10sinφ rdrdφ, on integrating with r = 0->5 and φ = 0->2π, we get Q = ψ = 0.
Answer: d
Explanation: Q = ∫ ρsds = ∫∫ 10sinφ rdrdφ, on integrating with r = 0->5 and φ = 0->2π, we
get Q = ψ = 0.
See lessThe total charge of a surface with densities 1,2,…,10 is
Answer: c Explanation: Q = ∫∫D.ds. Since the data is discrete, the total charge will be summation of 1,2,…,10,i.e, 1+2+…+10 = 10(11)/2 = 55.
Answer: c
Explanation: Q = ∫∫D.ds. Since the data is discrete, the total charge will be summation of
1,2,…,10,i.e, 1+2+…+10 = 10(11)/2 = 55.
See lessFind the electric field intensity of transformer oil (εr = 2 approx) with density 1/4π (in 109 units)
Answer: c Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units.
Answer: c
Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units.
See lessFor 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
Explanation: E = Q/ (4∏εor2
)
When distance d is infinity, the electric field will be zero, E= 0.
See lessThe 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
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.
See lessThe 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
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.
See lessFor a charge Q1, the effect of charge Q2 on Q1 will be,
Answer: b Explanation: The force of two charges with respect with each other is given by F1 and F2. Thus F1 + F2 = 0 and F1 = -F2.
Answer: b
Explanation: The force of two charges with respect with each other is given by F1 and
F2. Thus F1 + F2 = 0 and F1 = -F2.
See lessThe electric field intensity is defined as
Answer: c Explanation: The electric field intensity is the force per unit charge on a test charge, i.e, q1 = 1C. E = F/Q = Q/(4∏εr2).
Answer: c
Explanation: The electric field intensity is the force per unit charge on a test charge, i.e,
q1 = 1C. E = F/Q = Q/(4∏εr2).
See less