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What is the potential difference between 10sinθcosφ/r 2 at A(1,30,20) and B(4,90,60)?
Answer: c Explanation: Potential at A, Va = 10sin30cos20/12 = 4.6985 and Potential at B, Vb = 10sin90cos60/42 = 0.3125. Potential difference between A and B is, Vab = 4.6985 – 0.3125 = 4.386 volts.
Answer: c
Explanation: Potential at A, Va = 10sin30cos20/12 = 4.6985 and Potential at B, Vb =
10sin90cos60/42 = 0.3125. Potential difference between A and B is, Vab = 4.6985 –
0.3125 = 4.386 volts.
See lessThe voltage at any point in an ac circuit will be
Answer: b Explanation: In any ac circuit, the voltage measured will not be exact maximum. In order to normalise, we assume the instantaneous voltage at any point be 70.7% of the peak value, which is called the root mean square (RMS)voltage.
Answer: b
Explanation: In any ac circuit, the voltage measured will not be exact maximum. In order
to normalise, we assume the instantaneous voltage at any point be 70.7% of the peak
value, which is called the root mean square (RMS)voltage.
See lessDivergence theorem is based on
Answer: a Explanation: The divergence theorem relates surface integral and volume integral. Div(D) = ρv, which is Gauss’s law.
Answer: a
Explanation: The divergence theorem relates surface integral and volume integral.
Div(D) = ρv, which is Gauss’s law.
See lessThe 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 flux density of a sheet of charge density 25 units in air.
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.
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.
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 lessFind the potential of V = 60sin θ/r2 at P(3,60,25)
Answer: a Explanation: V = 60sin θ/r2 , put r = 3m, θ = 60 and φ = 25, V = 60 sin 60/32 = 5.774 volts.
Answer: a
Explanation: V = 60sin θ/r2
, put r = 3m, θ = 60 and φ = 25, V = 60 sin 60/32 = 5.774 volts.
See less